Heat developed in the heater is given by H = I2Rt Where, I = 15 A, R = 8 Ω and time t = 2 hours The rate at which heat is developed is given by 𝐻=𝐼²𝑅𝑡/𝑡=𝐼²𝑅=(15)²×8=1800 𝐽/𝑠 For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
Heat developed in the heater is given by H = I2Rt
Where, I = 15 A, R = 8 Ω and time t = 2 hours
The rate at which heat is developed is given by
𝐻=𝐼²𝑅𝑡/𝑡=𝐼²𝑅=(15)²×8=1800 𝐽/𝑠
Energy consumed by an electrical appliance is given by H = Pt For the TV set: Power P = 250 W and time t = 1 hour = 3600 seconds So, energy consumed H = 250 × 3600 = 900000 J For the toaster: Power P = 1200 W and time t = 10 minutes = 600 seconds So, energy consumed H = 1200 × 600 = 720000 J Hence,Read more
Energy consumed by an electrical appliance is given by H = Pt
For the TV set: Power P = 250 W and time t = 1 hour = 3600 seconds
So, energy consumed H = 250 × 3600 = 900000 J
For the toaster: Power P = 1200 W and time t = 10 minutes = 600 seconds
So, energy consumed H = 1200 × 600 = 720000 J
Hence, TV set uses more energy than toaster.
For the lamp one: Power P1 = 100 W and Potential Difference V = 220 V Therefore, 𝐼1=𝑃1/𝑉=100220=0.455 𝐴 For the lamp two: Power P2 = 60 W and Potential Difference V = 220 V Therefore, 𝐼2=𝑃2/𝑉=60/220=0.273 𝐴 So, the net current drawn from the supply is given by 𝐼=𝐼1+𝐼2=0.455+0.273=0.728 𝐴 For more anRead more
For the lamp one: Power P1 = 100 W and Potential Difference V = 220 V
Therefore,
𝐼1=𝑃1/𝑉=100220=0.455 𝐴
For the lamp two: Power P2 = 60 W and Potential Difference V = 220 V
Therefore,
𝐼2=𝑃2/𝑉=60/220=0.273 𝐴
So, the net current drawn from the supply is given by
𝐼=𝐼1+𝐼2=0.455+0.273=0.728 𝐴
Given that: Potential difference, V = 6 V (i) 1 Ω and 2 Ω resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω According to Ohm’s law, V = IR ⟹𝐼=𝑉/𝑅=6/3=2 𝐴 In series combination, the current in the circuit remains constant. Therefore power is given by 𝑃=𝐼²𝑅=(Read more
Given that: Potential difference, V = 6 V
(i) 1 Ω and 2 Ω resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law, V = IR
⟹𝐼=𝑉/𝑅=6/3=2 𝐴
In series combination, the current in the circuit remains constant. Therefore power is given by
𝑃=𝐼²𝑅=(2)²×2=8 𝑊
(ii) 1 Ω and 2 Ω resistors are connected in parallel. ⟹𝐼=𝑉/𝑅=6/3=2 𝐴
In parallel combination, the voltage in the circuit remains constant. Therefore, power is given by 𝑃=𝑉²/𝑅 =4²/2=8 𝑊
Hence, in both the cases power remains same as 8W.
Given that: Potential difference V = 220 V and resistance of each coil R = 24 Ω When the coil is used separately, the current in the coil is given by 𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴 When the two coils are connected in series, the net resistance is given by 𝑅=𝑅1+𝑅2=24 Ω+24 Ω=48 Ω Now, the current in the coilRead more
Given that: Potential difference V = 220 V and resistance of each coil R = 24 Ω
When the coil is used separately, the current in the coil is given by
𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴
When the two coils are connected in series, the net resistance is given by
𝑅=𝑅1+𝑅2=24 Ω+24 Ω=48 Ω
Now, the current in the coil is given by
𝐼=𝑉𝑅=220/48=55/12=4.58 𝐴
When the two coils are connected in parallel, the net resistance is given by 1/𝑅=1/24+1/24=2/24=1/12 ⟹𝑅=12 Ω
Now, the current in the coil is given by
𝐼=𝑉𝑅=22012=553=18.33 𝐴
For one bulb: Power P = 10 W and Potential Difference V = 220 V Using the relation for R, we have 𝑅 = 𝑉²/𝑃= (220) ²/10 = 4840 Ω Let the total number of bulbs be x. Given that: Current I = 5 A and Potential Difference V = 220 V According to Ohm’s law, V = IR ⟹𝑅=𝑉/𝐼=220/5=44 Ω Now, for x number of bulRead more
For one bulb: Power P = 10 W and Potential Difference V = 220 V
Using the relation for R, we have
𝑅 = 𝑉²/𝑃= (220) ²/10 = 4840 Ω
Let the total number of bulbs be x.
Given that: Current I = 5 A and Potential Difference V = 220 V
According to Ohm’s law, V = IR
⟹𝑅=𝑉/𝐼=220/5=44 Ω
Now, for x number of bulbs of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω.
1/44=1/4840 + 1/4840 + 1/4840 + ..𝑡𝑜 𝑥 𝑡𝑖𝑚𝑒𝑠 ⟹1/44=𝑥/4840 ⟹𝑥=484044=110
Therefore, 110 bulbs of 4840 Ω are required to draw the given amount of current.
Let the total number of resistors be x. Given that: Current I = 5 A and Potential Difference V = 220 V According to Ohm’s law, V = IR ⟹𝑅=𝑉/𝐼=220/5=44 Ω Now for x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω. 1/44= 1/176 +1/176 +1/1Read more
Let the total number of resistors be x.
Given that: Current I = 5 A and Potential Difference V = 220 V
According to Ohm’s law, V = IR
⟹𝑅=𝑉/𝐼=220/5=44 Ω
Now for x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω.
1/44= 1/176 +1/176 +1/176 + 1/176 +..𝑡𝑜 𝑥 𝑡𝑖𝑚𝑒𝑠 ⟹ 1/44=𝑥/176 ⟹𝑥=176/44=4
Therefore, 4 resistors of 176 Ω are required to draw the given amount of current.
Total resistance of resistors when connected in series is given by 𝑅=𝑅1+𝑅2+𝑅3+𝑅4+𝑅5 ⟹𝑅=0.2 Ω+0.3 Ω+0.4 Ω+0.5 Ω+12 Ω=13.4 Ω According to Ohm’s law, V = IR ⟹𝐼=𝑉/𝑅=9/13.4=0.67 𝐴 For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
Total resistance of resistors when connected in series is given by 𝑅=𝑅1+𝑅2+𝑅3+𝑅4+𝑅5 ⟹𝑅=0.2 Ω+0.3 Ω+0.4 Ω+0.5 Ω+12 Ω=13.4 Ω
According to Ohm’s law, V = IR
⟹𝐼=𝑉/𝑅=9/13.4=0.67 𝐴
According to Ohm’s law, V = IR ⟹𝑅=𝑉/𝐼 Here, V = 12 V and I = 2.5 mA = 0.0025 A Therefore, 𝑅=12/0.0025=4800 Ω=4.8 kΩ For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
According to Ohm’s law, V = IR
⟹𝑅=𝑉/𝐼
Here, V = 12 V and I = 2.5 mA = 0.0025 A
Therefore,
𝑅=12/0.0025=4800 Ω=4.8 kΩ
To measure the potential difference, a voltmeter should be connected in parallel. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
To measure the potential difference, a voltmeter should be connected in parallel.
An electric heater of resistance 8 W draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Heat developed in the heater is given by H = I2Rt Where, I = 15 A, R = 8 Ω and time t = 2 hours The rate at which heat is developed is given by 𝐻=𝐼²𝑅𝑡/𝑡=𝐼²𝑅=(15)²×8=1800 𝐽/𝑠 For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
Heat developed in the heater is given by H = I2Rt
Where, I = 15 A, R = 8 Ω and time t = 2 hours
The rate at which heat is developed is given by
𝐻=𝐼²𝑅𝑡/𝑡=𝐼²𝑅=(15)²×8=1800 𝐽/𝑠
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Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Energy consumed by an electrical appliance is given by H = Pt For the TV set: Power P = 250 W and time t = 1 hour = 3600 seconds So, energy consumed H = 250 × 3600 = 900000 J For the toaster: Power P = 1200 W and time t = 10 minutes = 600 seconds So, energy consumed H = 1200 × 600 = 720000 J Hence,Read more
Energy consumed by an electrical appliance is given by H = Pt
For the TV set: Power P = 250 W and time t = 1 hour = 3600 seconds
So, energy consumed H = 250 × 3600 = 900000 J
For the toaster: Power P = 1200 W and time t = 10 minutes = 600 seconds
So, energy consumed H = 1200 × 600 = 720000 J
Hence, TV set uses more energy than toaster.
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Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
For the lamp one: Power P1 = 100 W and Potential Difference V = 220 V Therefore, 𝐼1=𝑃1/𝑉=100220=0.455 𝐴 For the lamp two: Power P2 = 60 W and Potential Difference V = 220 V Therefore, 𝐼2=𝑃2/𝑉=60/220=0.273 𝐴 So, the net current drawn from the supply is given by 𝐼=𝐼1+𝐼2=0.455+0.273=0.728 𝐴 For more anRead more
For the lamp one: Power P1 = 100 W and Potential Difference V = 220 V
Therefore,
𝐼1=𝑃1/𝑉=100220=0.455 𝐴
For the lamp two: Power P2 = 60 W and Potential Difference V = 220 V
Therefore,
𝐼2=𝑃2/𝑉=60/220=0.273 𝐴
So, the net current drawn from the supply is given by
𝐼=𝐼1+𝐼2=0.455+0.273=0.728 𝐴
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Compare the power used in the 2 ohm resistor in each of the following circuits:
Given that: Potential difference, V = 6 V (i) 1 Ω and 2 Ω resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω According to Ohm’s law, V = IR ⟹𝐼=𝑉/𝑅=6/3=2 𝐴 In series combination, the current in the circuit remains constant. Therefore power is given by 𝑃=𝐼²𝑅=(Read more
Given that: Potential difference, V = 6 V
(i) 1 Ω and 2 Ω resistors are connected in series. So, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law, V = IR
⟹𝐼=𝑉/𝑅=6/3=2 𝐴
In series combination, the current in the circuit remains constant. Therefore power is given by
𝑃=𝐼²𝑅=(2)²×2=8 𝑊
(ii) 1 Ω and 2 Ω resistors are connected in parallel. ⟹𝐼=𝑉/𝑅=6/3=2 𝐴
In parallel combination, the voltage in the circuit remains constant. Therefore, power is given by 𝑃=𝑉²/𝑅 =4²/2=8 𝑊
Hence, in both the cases power remains same as 8W.
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A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 ohm resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Given that: Potential difference V = 220 V and resistance of each coil R = 24 Ω When the coil is used separately, the current in the coil is given by 𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴 When the two coils are connected in series, the net resistance is given by 𝑅=𝑅1+𝑅2=24 Ω+24 Ω=48 Ω Now, the current in the coilRead more
Given that: Potential difference V = 220 V and resistance of each coil R = 24 Ω
When the coil is used separately, the current in the coil is given by
𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴
When the two coils are connected in series, the net resistance is given by
𝑅=𝑅1+𝑅2=24 Ω+24 Ω=48 Ω
Now, the current in the coil is given by
𝐼=𝑉𝑅=220/48=55/12=4.58 𝐴
When the two coils are connected in parallel, the net resistance is given by 1/𝑅=1/24+1/24=2/24=1/12 ⟹𝑅=12 Ω
Now, the current in the coil is given by
𝐼=𝑉𝑅=22012=553=18.33 𝐴
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Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
For one bulb: Power P = 10 W and Potential Difference V = 220 V Using the relation for R, we have 𝑅 = 𝑉²/𝑃= (220) ²/10 = 4840 Ω Let the total number of bulbs be x. Given that: Current I = 5 A and Potential Difference V = 220 V According to Ohm’s law, V = IR ⟹𝑅=𝑉/𝐼=220/5=44 Ω Now, for x number of bulRead more
For one bulb: Power P = 10 W and Potential Difference V = 220 V
Using the relation for R, we have
𝑅 = 𝑉²/𝑃= (220) ²/10 = 4840 Ω
Let the total number of bulbs be x.
Given that: Current I = 5 A and Potential Difference V = 220 V
According to Ohm’s law, V = IR
⟹𝑅=𝑉/𝐼=220/5=44 Ω
Now, for x number of bulbs of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω.
1/44=1/4840 + 1/4840 + 1/4840 + ..𝑡𝑜 𝑥 𝑡𝑖𝑚𝑒𝑠 ⟹1/44=𝑥/4840 ⟹𝑥=484044=110
Therefore, 110 bulbs of 4840 Ω are required to draw the given amount of current.
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How many 176 ohm resistors (in parallel) are required to carry 5 A on a 220 V line?
Let the total number of resistors be x. Given that: Current I = 5 A and Potential Difference V = 220 V According to Ohm’s law, V = IR ⟹𝑅=𝑉/𝐼=220/5=44 Ω Now for x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω. 1/44= 1/176 +1/176 +1/1Read more
Let the total number of resistors be x.
Given that: Current I = 5 A and Potential Difference V = 220 V
According to Ohm’s law, V = IR
⟹𝑅=𝑉/𝐼=220/5=44 Ω
Now for x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω.
1/44= 1/176 +1/176 +1/176 + 1/176 +..𝑡𝑜 𝑥 𝑡𝑖𝑚𝑒𝑠 ⟹ 1/44=𝑥/176 ⟹𝑥=176/44=4
Therefore, 4 resistors of 176 Ω are required to draw the given amount of current.
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A battery of 9 V is connected in series with resistors of 0.2 ohm, 0.3 ohm, 0.4 ohm , 0.5 ohm and 12 ohm, respectively. How much current would flow through the 12 ohm resistor?
Total resistance of resistors when connected in series is given by 𝑅=𝑅1+𝑅2+𝑅3+𝑅4+𝑅5 ⟹𝑅=0.2 Ω+0.3 Ω+0.4 Ω+0.5 Ω+12 Ω=13.4 Ω According to Ohm’s law, V = IR ⟹𝐼=𝑉/𝑅=9/13.4=0.67 𝐴 For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
Total resistance of resistors when connected in series is given by 𝑅=𝑅1+𝑅2+𝑅3+𝑅4+𝑅5 ⟹𝑅=0.2 Ω+0.3 Ω+0.4 Ω+0.5 Ω+12 Ω=13.4 Ω
According to Ohm’s law, V = IR
⟹𝐼=𝑉/𝑅=9/13.4=0.67 𝐴
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When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
According to Ohm’s law, V = IR ⟹𝑅=𝑉/𝐼 Here, V = 12 V and I = 2.5 mA = 0.0025 A Therefore, 𝑅=12/0.0025=4800 Ω=4.8 kΩ For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
According to Ohm’s law, V = IR
⟹𝑅=𝑉/𝐼
Here, V = 12 V and I = 2.5 mA = 0.0025 A
Therefore,
𝑅=12/0.0025=4800 Ω=4.8 kΩ
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How is a voltmeter connected in the circuit to measure the potential difference between two points?
To measure the potential difference, a voltmeter should be connected in parallel. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
To measure the potential difference, a voltmeter should be connected in parallel.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/