Current Electricity

JEE Main level, single correct, medium

206 of 380

Question 206 of 380

The voltage and the current between AA and BB points shown in the circuit are _____.

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Answer: (B) 2424 V, 44 A

The four parallel branches between the common left node AA and the common right node are all identical: each consists of a 27 V27\text{ V} source in series with a 3 Ω3\ \Omega resistor. In the third branch, the two cells (14 V(14\text{ V} and 13 V)13\text{ V}) aid each other, giving a net EMF of 14+13=27 V14 + 13 = 27\text{ V}.

Thevenin equivalent between
AA and BB:
Since all four branches have the same EMF and polarity in parallel:

Vth=27 VV_{th} = 27\text{ V}

The four
3 Ω3\ \Omega resistors in parallel give:
Rth=34=0.75 ΩR_{th} = \dfrac{3}{4} = 0.75\ \Omega

Load branch (diagonal):

The diagonal connecting
AA and BB consists of two 3 Ω3\ \Omega resistors in series, so:
RL=3+3=6 ΩR_L = 3 + 3 = 6\ \Omega

Current through the diagonal branch:

I=VthRth+RL=270.75+6=276.75=4 AI = \dfrac{V_{th}}{R_{th} + R_L} = \dfrac{27}{0.75 + 6} = \dfrac{27}{6.75} = 4\text{ A}

Voltage across
AA and BB:
VAB=I⋅RL=4×6=24 VV_{AB} = I \cdot R_L = 4 \times 6 = 24\text{ V}

Hence, the voltage is
24 V24\text{ V} and the current is 4 A4\text{ A}, which corresponds to option (2) 24 V, 4 A(2)\ 24\text{ V},\ 4\text{ A}.

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