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Question

In the electric circuit shown in the figure, current flowing through 4 Ω resistor is :

79963_4a4d5a9ff8454042b8b0597a553e5798.png

A
3A.
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B
2A
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C
1A
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D
4A
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Solution

The correct option is C 1A
The circuit can be reduced, step by step, to a single equivalent resistance. The 8 ohm and the 8 ohm are connected in parallel , and so they can be replaced by an equivalent resistor Rp of 4 ohms, using the below equation.

Rp=(8×8)(8+8)=4 ohms

This resistor is connected in series with the 4 ohm resistor R1. The total resistance Rt of the circuit is then

Rt=R1+Rp=4+4=8 ohms.

Since the 4 ohm and the 4 ohm are connected in series, they have the same current I1, which must be equal to the current of the battery. Using Ohms law we get,

I=VR=88=1A.

Hence, the current flowing through the resistor R1 (4 ohms) is 1A.

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