Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω Potential difference, V = 9 V Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A ∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.
Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω
Potential difference, V = 9 V
Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A
∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6/2= 3ohm which is also not desired. Hence, we should eitRead more
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be
6/2= 3ohm
which is also not desired. Hence, we should either connect the two resistors in series or parallel.
(i) Two resistors in parallel
Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be
1/(1/6 + 1/6) = 3ohm
The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.
(ii) Two resistors in series
Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω
The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be
Here, current, I = 5 A, voltage, V = 220 V ∴ Maxium power, P = I x V = 5 x 220 = 1100W Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110 ∴ 110 lamps can be connected in parallel.
Here, current, I = 5 A, voltage, V = 220 V
∴ Maxium power, P = I x V = 5 x 220 = 1100W
Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110
∴ 110 lamps can be connected in parallel.
Supply voltage, V = 220 V Resistance of one coil, R = 24 Ω (i) Coils are used separately According to Ohm’s law, V = I1R1 Where, I1 is the current flowing through the coil 𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴 Therefore, 9.16 A current will flow through the coil when used separately. (ii) Coils are connectRead more
Supply voltage, V = 220 V
Resistance of one coil, R = 24 Ω
(i) Coils are used separately
According to Ohm’s law,
V = I1R1
Where,
I1 is the current flowing through the coil
𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series
Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω
According to Ohm’s law,
V = I2R2
Where,
I2 is the current flowing through the series circuit
𝐼=𝑉𝑅=220/48=55/12=4.58 𝐴
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel
Total resistance, R3 is given as
1/𝑅=1/24+1/24=2/24=1/12 ⟹𝑅=12 Ω
According to Ohm’s law,
V = I3R3
Where,
I3 is the current flowing through the circuit
𝐼=𝑉𝑅=22012=553=18.33 𝐴
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.
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.
Here, V = 12 V and I = 2.5 mA = 2.5 x 10-3 A ∴ Resistance, R = V/I = 12V/(2.5× 103A) = 4,800 Ω = 4.8 x 10-3 Ω
Here, V = 12 V and I = 2.5 mA = 2.5 x 10-3 A
See less∴ Resistance, R = V/I = 12V/(2.5× 103A) = 4,800 Ω = 4.8 x 10-3 Ω
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, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω Potential difference, V = 9 V Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A ∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.
Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω
See lessPotential difference, V = 9 V
Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A
∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.
Show how you would connect three resistors, each of resistance 6 W, so that the combination has a resistance of (i) 9 ohm, (ii) 4 ohm.
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6/2= 3ohm which is also not desired. Hence, we should eitRead more
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be
6/2= 3ohm
which is also not desired. Hence, we should either connect the two resistors in series or parallel.
(i) Two resistors in parallel
Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be
1/(1/6 + 1/6) = 3ohm
The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.
(ii) Two resistors in series
Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω
The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be
1/(1/12 + 1/6) = 4ohm
Therefore, the total resistance is 4 Ω.
See lessSeveral 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?
Here, current, I = 5 A, voltage, V = 220 V ∴ Maxium power, P = I x V = 5 x 220 = 1100W Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110 ∴ 110 lamps can be connected in parallel.
Here, current, I = 5 A, voltage, V = 220 V
See less∴ Maxium power, P = I x V = 5 x 220 = 1100W
Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110
∴ 110 lamps can be connected in parallel.
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?
Supply voltage, V = 220 V Resistance of one coil, R = 24 Ω (i) Coils are used separately According to Ohm’s law, V = I1R1 Where, I1 is the current flowing through the coil 𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴 Therefore, 9.16 A current will flow through the coil when used separately. (ii) Coils are connectRead more
Supply voltage, V = 220 V
Resistance of one coil, R = 24 Ω
(i) Coils are used separately
According to Ohm’s law,
V = I1R1
Where,
I1 is the current flowing through the coil
𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series
Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω
According to Ohm’s law,
V = I2R2
Where,
I2 is the current flowing through the series circuit
𝐼=𝑉𝑅=220/48=55/12=4.58 𝐴
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel
Total resistance, R3 is given as
1/𝑅=1/24+1/24=2/24=1/12 ⟹𝑅=12 Ω
According to Ohm’s law,
V = I3R3
Where,
I3 is the current flowing through the circuit
𝐼=𝑉𝑅=22012=553=18.33 𝐴
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.
See less