Current Electricity

JEE Main level, single correct, medium

94 of 380

Question 94 of 380

Refer to the figure given below, current between terminals AA and BB is _____ A.

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Answer: (B) 1.251.25

The given circuit consists of four parallel branches connected between the left and right common vertical wires.

Let's find the equivalent electromotive force (EMF) and resistance for each branch:
For the top three active branches, each contains three 5 V5\text{ V} cells and three 3 Ω3\ \Omega resistors in series.
Equivalent EMF of each active branch, E=5+5+5=15 VE = 5 + 5 + 5 = 15\text{ V}.
Equivalent resistance of each active branch, R=3+3+3=9 ΩR = 3 + 3 + 3 = 9\ \Omega.

Since these three identical active branches are connected in parallel, we can replace them with a single equivalent voltage source:
Equivalent EMF, Eeq=15 VE_{eq} = 15\text{ V} (as they are identical and in parallel).
Equivalent internal resistance, Req=93=3 ΩR_{eq} = \dfrac{9}{3} = 3\ \Omega.

The fourth (bottom) branch is connected between terminals AA and BB and acts as the load for this equivalent source. It contains three 3 Ω3\ \Omega resistors in series.
Resistance of the branch ABAB, RAB=3+3+3=9 ΩR_{AB} = 3 + 3 + 3 = 9\ \Omega.

Now, the circuit is simplified to an equivalent source of 15 V15\text{ V} with internal resistance 3 Ω3\ \Omega connected across a load of 9 Ω9\ \Omega.
The current II flowing between terminals AA and BB (through the bottom branch) is given by Ohm's law:
I=EeqReq+RABI = \dfrac{E_{eq}}{R_{eq} + R_{AB}}

I=153+9=1512 AI = \dfrac{15}{3 + 9} = \dfrac{15}{12}\text{ A}

I=1.25 AI = 1.25\text{ A}

Answer: 1.251.25

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