The current in a simple series circuit is 5-0A- When an additional resistance of 2-0- is inserted- the current drops to 4-0A- The original resistance of the circuit in ohms was-

The correct option is C 8 Given : #160; I_1 = 5 A #160; #160; #160; #160; #160; #160; #160; R' = 2Ω #160; #160; ...

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Standard XII

Physics

Mixed Combination of Cells

The current i...

Question

The current in a simple series circuit is $5.0$ A. When an additional resistance of $2.0 Ω$ is inserted, the current drops to $4.0$ A. The original resistance of the circuit in ohms was.

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Solution

The correct option is C $8$
Given : $I_{1} = 5$ A $R^{'} = 2 Ω$ $I_{2} = 4$ A
Let the original resistance of the circuit be $R$ .
Voltage given by the battery is same in both the circuits i.e $V_{1} = V_{2}$
$∴$ $I_{1} R = I_{2} (R + R^{'})$
$∴$ $5 R = 4 (R + 2)$ $⟹ R = 8$ ohm

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The current in a simple series circuit is 5.0A. When an additional resistance of 2.0Ω is inserted, the current drops to 4.0A. The original resistance of the circuit in ohms was.

The correct option is C 8Given : I1=5 A R′=2Ω I2=4 ALet the original resistance of the circuit be R.Voltage given by the battery is same in both the circuits i.e V1=V2∴ I1R=I2(R+R′)∴ 5R=4(R+2) ⟹R=8 ohm

The current in a simple series circuit is $5.0$ A. When an additional resistance of $2.0 Ω$ is inserted, the current drops to $4.0$ A. The original resistance of the circuit in ohms was.

The correct option is C $8$
Given : $I_{1} = 5$ A $R^{'} = 2 Ω$ $I_{2} = 4$ A
Let the original resistance of the circuit be $R$ .
Voltage given by the battery is same in both the circuits i.e $V_{1} = V_{2}$
$∴$ $I_{1} R = I_{2} (R + R^{'})$
$∴$ $5 R = 4 (R + 2)$ $⟹ R = 8$ ohm