In the circuit shown, the current in the 1 Ω resistor is: (a) 0.13 A, from Q to P (b) 0.13 A, from P to Q (c) 1.3 A from P to Q (d) 0 A

Answer: (a). Take VQ = 0, VP = x. Left branch: 6 V cell with 3 Ω; right branch: 9 V cell (+ at top) with 2 + 3 = 5 Ω; middle: 1 Ω. Left branch current Q→P: -(x + 6)/(3); right branch current P→Q: (x - 9)/(5); 1 Ω current P→Q: x. KCL at P: -x+6/3 = x + x-9/5 ⇒ -23x = 3 x = -0.13 V, so 0.13 A flows through 1 Ω from Q to P. So the answer is (a) 0.13 A, from Q to P.

In the circuit shown, the current in the 1 Ω resistor is: | JEE Main 2015

Q179JEE Main 2015Medium

Kirchhoff's Laws & Measuring Instruments

In the circuit shown, the current in the $1\ \Omega$ resistor is:

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Answer: (a)

Take $V_Q = 0$ , $V_P = x$ . Left branch: $6$ V cell with $3\ \Omega$ ; right branch: $9$ V cell (+ at top) with $2 + 3 = 5\ \Omega$ ; middle: $1\ \Omega$ .
Left branch current Q→P: $-\frac{x + 6}{3}$ ; right branch current P→Q: $\frac{x - 9}{5}$ ; $1\ \Omega$ current P→Q: $x$ .
KCL at P: $-\frac{x+6}{3} = x + \frac{x-9}{5} \Rightarrow -23x = 3$
$x = -0.13$ V, so $0.13$ A flows through $1\ \Omega$ from Q to P.
So the answer is (a) $0.13$ A, from Q to P.

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