Given, Electric current (I) = 5A, Potential difference (V) = 220V, Time (t) = 2h = 2 x 60 x 60 s = 7200 s Power (P) =? Energy consumed =? We know that, P = VI = 220V × 5A = 1100W We know that energy consumed by the electric appliance = P × t ⇒ Energy consumed = 1100 W × 7200 s = 7920000 J = 7.92 × 1Read more
Given, Electric current (I) = 5A,
Potential difference (V) = 220V,
Time (t) = 2h = 2 x 60 x 60 s = 7200 s
Power (P) =?
Energy consumed =?
We know that, P = VI = 220V × 5A = 1100W
We know that energy consumed by the electric appliance = P × t ⇒ Energy consumed = 1100 W × 7200 s = 7920000 J = 7.92 × 106 J
Since electric power is the rate of consumption of electric energy in any electrical appliance. Hence, rate at which energy is delivered by a current is the power of electric appliance.
Since electric power is the rate of consumption of electric energy in any electrical appliance. Hence, rate at which energy is delivered by a current is the power of electric appliance.
Given, Electric current (I) = 5A, Resistance (R) = 20 Ω, Time (t) = 30s Therefore, Heat produced (H) =? Since, V=I×R=5A×20Ω=100V We know that, Heat produced (H) = VIt ⇒H=100V×5A×30s=15000J=1.5×104J
Given, Electric current (I) = 5A, Resistance (R) = 20 Ω, Time (t) = 30s
Therefore, Heat produced (H) =?
Since, V=I×R=5A×20Ω=100V
We know that, Heat produced (H) = VIt ⇒H=100V×5A×30s=15000J=1.5×104J
Given, Potential difference (V) = 50V, Charge (Q) = 96000 coulomb Time, t = 1 hour = 60 X 60 s = 3600 s Heat produced =? We know that electric current (I) = Q/t=96000C/3600s=96000/3600A We know that heat produced (H) in the given time (t) = VIt H=50V×(96000/3600A)×3600=50×96000J=4800000J=4.8×106 JouRead more
Given, Potential difference (V) = 50V,
Charge (Q) = 96000 coulomb
Time, t = 1 hour = 60 X 60 s = 3600 s
Heat produced =?
We know that electric current (I) = Q/t=96000C/3600s=96000/3600A
We know that heat produced (H) in the given time (t) = VIt H=50V×(96000/3600A)×3600=50×96000J=4800000J=4.8×106 Joule
The resistance of heating element is very high compared to the cord of an electric heater. Since, heat produced is directly proportional to the resistance, thus element of electric heater glows because of production of more heat and cord does not.
The resistance of heating element is very high compared to the cord of an electric heater. Since, heat produced is directly proportional to the resistance, thus element of electric heater glows because of production of more heat and cord does not.
(a) When all the resistors are connected in parallel Then the 1/R=1/4Ω+1/8Ω+1/12Ω+1/24Ω=(6+3+2+1)/24Ω=12/24Ω=1/2Ω Thus, R = 2 Ω (b) When all the resistors are connected in series Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω Thus, highest resistance = 48 Ω Lowest resistance = 2 Ω
(a) When all the resistors are connected in parallel
Then the 1/R=1/4Ω+1/8Ω+1/12Ω+1/24Ω=(6+3+2+1)/24Ω=12/24Ω=1/2Ω
Thus, R = 2 Ω
(b) When all the resistors are connected in series
Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω
Thus, highest resistance = 48 Ω
Lowest resistance = 2 Ω
(a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1 Therefore, 1/R1=1/6Ω+1/3Ω=1+2/6Ω=3/6Ω=1/2Ω Thus R1=2Ω Now, total effectiRead more
(a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series
Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1 Therefore, 1/R1=1/6Ω+1/3Ω=1+2/6Ω=3/6Ω=1/2Ω
Thus R1=2Ω
Now, total effective resistance in the circuit = R1 + 2 Ω = 2 Ω + 2 Ω = 4 Ω
Hence, when resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series, then the total effective resistance in the circuit = 4 Ω
(b) When all the three resistance is connected in parallel then 1/R=1/2Ω+1/3Ω+1/6Ω=3+2+1/6Ω=6/6Ω=1Ω
Thus, R=1Ω
When all the three resistance will be connected in parallel, then the total effective resistance in the circuit = 1 Ω
Advantages of connecting electrical appliances in parallel instead of connecting in series: (a) Voltage remains same in all the appliances. (b) Lower total effective resistance
Advantages of connecting electrical appliances in parallel instead of connecting in series:
(a) Voltage remains same in all the appliances.
(b) Lower total effective resistance
Given: R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel Potential difference = 220V Thus, 1/R=1/100Ω+1/50Ω+1/500Ω=(5+10+1)/500Ω=16/500Ω Therefore, R=500/16Ω=31.25Ω Electric current (I) through the circuit = V/R ⇒I=220V/31.25Ω=7.04A For electric iron Since it takes as well current as thRead more
Given:
R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel
Potential difference = 220V
Thus, 1/R=1/100Ω+1/50Ω+1/500Ω=(5+10+1)/500Ω=16/500Ω
Therefore, R=500/16Ω=31.25Ω
Electric current (I) through the circuit = V/R ⇒I=220V/31.25Ω=7.04A
For electric iron
Since it takes as well current as three appliances, thus electric current through it = 7.04A
The electric current = 7.04 A and potential difference = 220 V
Thus, Resistance of electric iron = Total resistance of three appliances = 31.25 Ω
Thus, electric current through the electric iron = 7.04A
Resistance of electric iron = 31.25 Ω
Since 1/R=1/R1+1/R2+1/R3+..+1/Rn when resistors are connected in parallel (a) 1 Ω and 106 Ω Thus, 1/R=1/1Ω+1/106Ω=106+1/106Ω=107/106Ω Thus, R=106/107Ω=0.99Ω Thus, equivalent resistance of 1Ω and 106Ω are connected in parallel = 0.99Ω (b) 1 Ω and 103 Ω, and 106 Ω Thus, 1/R=1/1Ω+1/103Ω+1/106Ω=(10918+Read more
Since 1/R=1/R1+1/R2+1/R3+..+1/Rn
when resistors are connected in parallel
(a) 1 Ω and 106 Ω
Thus, 1/R=1/1Ω+1/106Ω=106+1/106Ω=107/106Ω
Thus, R=106/107Ω=0.99Ω
Thus, equivalent resistance of 1Ω and 106Ω are connected in parallel = 0.99Ω
(b) 1 Ω and 103 Ω, and 106 Ω
Thus, 1/R=1/1Ω+1/103Ω+1/106Ω=(10918+106+103)/103×106Ω=11127/10918Ω
Thus, R=10918/11127Ω=1.02Ω
Thus, equivalent resistance of 1 Ω, 103 Ω and 106 Ω are connected in parallel = 1.02Ω
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Given, Electric current (I) = 5A, Potential difference (V) = 220V, Time (t) = 2h = 2 x 60 x 60 s = 7200 s Power (P) =? Energy consumed =? We know that, P = VI = 220V × 5A = 1100W We know that energy consumed by the electric appliance = P × t ⇒ Energy consumed = 1100 W × 7200 s = 7920000 J = 7.92 × 1Read more
Given, Electric current (I) = 5A,
Potential difference (V) = 220V,
See lessTime (t) = 2h = 2 x 60 x 60 s = 7200 s
Power (P) =?
Energy consumed =?
We know that, P = VI = 220V × 5A = 1100W
We know that energy consumed by the electric appliance = P × t
⇒ Energy consumed = 1100 W × 7200 s = 7920000 J = 7.92 × 106 J
What determines the rate at which energy is delivered by a current?
Since electric power is the rate of consumption of electric energy in any electrical appliance. Hence, rate at which energy is delivered by a current is the power of electric appliance.
Since electric power is the rate of consumption of electric energy in any electrical appliance. Hence, rate at which energy is delivered by a current is the power of electric appliance.
See lessAn electric iron of resistance 20 ohm takes a current of 5 A. Calculate the heat developed in 30 s.
Given, Electric current (I) = 5A, Resistance (R) = 20 Ω, Time (t) = 30s Therefore, Heat produced (H) =? Since, V=I×R=5A×20Ω=100V We know that, Heat produced (H) = VIt ⇒H=100V×5A×30s=15000J=1.5×104J
Given, Electric current (I) = 5A, Resistance (R) = 20 Ω, Time (t) = 30s
See lessTherefore, Heat produced (H) =?
Since, V=I×R=5A×20Ω=100V
We know that, Heat produced (H) = VIt
⇒H=100V×5A×30s=15000J=1.5×104J
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Given, Potential difference (V) = 50V, Charge (Q) = 96000 coulomb Time, t = 1 hour = 60 X 60 s = 3600 s Heat produced =? We know that electric current (I) = Q/t=96000C/3600s=96000/3600A We know that heat produced (H) in the given time (t) = VIt H=50V×(96000/3600A)×3600=50×96000J=4800000J=4.8×106 JouRead more
Given, Potential difference (V) = 50V,
See lessCharge (Q) = 96000 coulomb
Time, t = 1 hour = 60 X 60 s = 3600 s
Heat produced =?
We know that electric current (I) = Q/t=96000C/3600s=96000/3600A
We know that heat produced (H) in the given time (t) = VIt
H=50V×(96000/3600A)×3600=50×96000J=4800000J=4.8×106 Joule
Why does the cord of an electric heater not glow while the heating element does?
The resistance of heating element is very high compared to the cord of an electric heater. Since, heat produced is directly proportional to the resistance, thus element of electric heater glows because of production of more heat and cord does not.
The resistance of heating element is very high compared to the cord of an electric heater. Since, heat produced is directly proportional to the resistance, thus element of electric heater glows because of production of more heat and cord does not.
See lessWhat is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 ohm, 8 ohm, 12 ohm, 24 ohm?
(a) When all the resistors are connected in parallel Then the 1/R=1/4Ω+1/8Ω+1/12Ω+1/24Ω=(6+3+2+1)/24Ω=12/24Ω=1/2Ω Thus, R = 2 Ω (b) When all the resistors are connected in series Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω Thus, highest resistance = 48 Ω Lowest resistance = 2 Ω
(a) When all the resistors are connected in parallel
Then the 1/R=1/4Ω+1/8Ω+1/12Ω+1/24Ω=(6+3+2+1)/24Ω=12/24Ω=1/2Ω
Thus, R = 2 Ω
(b) When all the resistors are connected in series
See lessThus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω
Thus, highest resistance = 48 Ω
Lowest resistance = 2 Ω
How can three resistors of resistances 2 ohm, 3 ohm, and 6 ohm be connected to give a total resistance of (a) 4 ohm, (b) 1 ohm?
(a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1 Therefore, 1/R1=1/6Ω+1/3Ω=1+2/6Ω=3/6Ω=1/2Ω Thus R1=2Ω Now, total effectiRead more
(a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series
Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1
Therefore, 1/R1=1/6Ω+1/3Ω=1+2/6Ω=3/6Ω=1/2Ω
Thus R1=2Ω
Now, total effective resistance in the circuit = R1 + 2 Ω = 2 Ω + 2 Ω = 4 Ω
Hence, when resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series, then the total effective resistance in the circuit = 4 Ω
(b) When all the three resistance is connected in parallel then
See less1/R=1/2Ω+1/3Ω+1/6Ω=3+2+1/6Ω=6/6Ω=1Ω
Thus, R=1Ω
When all the three resistance will be connected in parallel, then the total effective resistance in the circuit = 1 Ω
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Advantages of connecting electrical appliances in parallel instead of connecting in series: (a) Voltage remains same in all the appliances. (b) Lower total effective resistance
Advantages of connecting electrical appliances in parallel instead of connecting in series:
See less(a) Voltage remains same in all the appliances.
(b) Lower total effective resistance
An electric lamp of 100 ohm, a toaster of resistance 50 ohm, and a water filter of resistance 500 ohm are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Given: R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel Potential difference = 220V Thus, 1/R=1/100Ω+1/50Ω+1/500Ω=(5+10+1)/500Ω=16/500Ω Therefore, R=500/16Ω=31.25Ω Electric current (I) through the circuit = V/R ⇒I=220V/31.25Ω=7.04A For electric iron Since it takes as well current as thRead more
Given:
See lessR1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel
Potential difference = 220V
Thus, 1/R=1/100Ω+1/50Ω+1/500Ω=(5+10+1)/500Ω=16/500Ω
Therefore, R=500/16Ω=31.25Ω
Electric current (I) through the circuit = V/R
⇒I=220V/31.25Ω=7.04A
For electric iron
Since it takes as well current as three appliances, thus electric current through it = 7.04A
The electric current = 7.04 A and potential difference = 220 V
Thus, Resistance of electric iron = Total resistance of three appliances = 31.25 Ω
Thus, electric current through the electric iron = 7.04A
Resistance of electric iron = 31.25 Ω
Judge the equivalent resistance when the following are connected in parallel
Since 1/R=1/R1+1/R2+1/R3+..+1/Rn when resistors are connected in parallel (a) 1 Ω and 106 Ω Thus, 1/R=1/1Ω+1/106Ω=106+1/106Ω=107/106Ω Thus, R=106/107Ω=0.99Ω Thus, equivalent resistance of 1Ω and 106Ω are connected in parallel = 0.99Ω (b) 1 Ω and 103 Ω, and 106 Ω Thus, 1/R=1/1Ω+1/103Ω+1/106Ω=(10918+Read more
Since 1/R=1/R1+1/R2+1/R3+..+1/Rn
when resistors are connected in parallel
(a) 1 Ω and 106 Ω
Thus, 1/R=1/1Ω+1/106Ω=106+1/106Ω=107/106Ω
Thus, R=106/107Ω=0.99Ω
Thus, equivalent resistance of 1Ω and 106Ω are connected in parallel = 0.99Ω
(b) 1 Ω and 103 Ω, and 106 Ω
See lessThus, 1/R=1/1Ω+1/103Ω+1/106Ω=(10918+106+103)/103×106Ω=11127/10918Ω
Thus, R=10918/11127Ω=1.02Ω
Thus, equivalent resistance of 1 Ω, 103 Ω and 106 Ω are connected in parallel = 1.02Ω