JEE Main 2014 · 12 Apr · Shift 1 · Physics Q18

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PhysicsSingle correct+4 / -1Question 18 of 90

The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 Ω, 9 Ω and 9 Ω and a capacitor 5.0 μF.

How much is the current I in the circuit in steady state ?

Three vertical branches between a top and a bottom wire: left, 3 Ω in series with the 8.0 V battery; middle, the 5 μF capacitor above a 9 Ω resistor; right, 9 Ω in series with the 16.0 V battery. Current I is marked on the top wire (flowing left) and I₁, I₂ on the bottom wire (flowing right).

1.6 A

0.67 A

2.5 A

0.25 A

Answer and worked solution

Official answer: B

In steady state no current flows through the capacitor branch. The 8 V and 16 V batteries oppose round the outer loop through 3 Ω and 9 Ω:
$I=\frac{16-8}{3+9}=0.67\text{ A}.$

Topics:Kirchhoff's laws RC circuits

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 18