The current flowing through the 1 Ω resistor is n/10 A. The value of n is _______. Numerical answer type. Work it out, then open the answer.

Answer: 25. Take VA = 0. The 5 V cell gives VB - VD = 5; current in the 1 Ω branch C to A is I₁ = VC + 10. KCL for the B+D pair: (VB + VD)/(4) = (2VC - VB - VD)/(2) ⇒ VC = 3/4(VB + VD) KCL at C: (VB + VD - 2VC)/(2) = VC + 10 ⇒ VB + VD = -10, VC = -7.5 V I₁ = -7.5 + 10 = 2.5 A = 25/10 A So n = 25.

The current flowing through the 1 Ω resistor is n/10 A. The value of… | JEE Main 2024 (9 Apr, Shift 1)

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Kirchhoff's Laws & Measuring Instruments

The current flowing through the $1\ \Omega$ resistor is $\frac{n}{10}$ A. The value of $n$ is _______.

Numerical answer type. Work it out, then open the answer.

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Answer: 25

Take $V_A = 0$ . The 5 V cell gives $V_B - V_D = 5$ ; current in the $1\ \Omega$ branch C to A is $I_1 = V_C + 10$ .
KCL for the B+D pair: $\frac{V_B + V_D}{4} = \frac{2V_C - V_B - V_D}{2} \Rightarrow V_C = \frac{3}{4}(V_B + V_D)$
KCL at C: $\frac{V_B + V_D - 2V_C}{2} = V_C + 10 \Rightarrow V_B + V_D = -10$ , $V_C = -7.5\text{ V}$
$I_1 = -7.5 + 10 = 2.5\text{ A} = \frac{25}{10}\text{ A}$
So $n = 25$ .

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