When an ac generator of 120 V is connected in series with a capacitor and a resistor of 30 - the circuit carries a current 1-5 A- The potential difference across the capacitor will be

Voltage across resistance nbsp; = IR = 1.5 × 30 = 45 V And in RC circuit voltage drop is given by V^2 = V^2_R + V^2_C V_C = nbsp;√(V^2 V_R^2) = √((120) ...

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Byju's Answer

Standard XII

Physics

Kirchhoff's Voltage Law

When an ac ge...

Question

When an ac generator of $120 V$ is connected in series with a capacitor and a resistor of $30 Ω$ , the circuit carries a current $1.5 A$ . The potential difference across the capacitor will be

1. 11 V

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111 V

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75 V

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Zero

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Solution

The correct option is B $111 V$
Voltage across resistance $= I R = 1.5 \times 30 = 45 V$

And in $R C$ circuit voltage drop is given by

$V^{2} = V_{R}^{2} + V_{C}^{2}$

$V_{C} = \sqrt{V^{2} - V_{R}^{2}} = \sqrt{(120)^{2} - (45)^{2}} = 111 V$

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When an ac generator of 120 V is connected in series with a capacitor and a resistor of 30 Ω, the circuit carries a current 1.5 A. The potential difference across the capacitor will be

The correct option is B 111 VVoltage across resistance =IR=1.5×30=45 V And in RC circuit voltage drop is given by V2=V2R+V2C VC=√V2−V2R=√(120)2−(45)2=111 V

When an ac generator of $120 V$ is connected in series with a capacitor and a resistor of $30 Ω$ , the circuit carries a current $1.5 A$ . The potential difference across the capacitor will be

The correct option is B $111 V$
Voltage across resistance $= I R = 1.5 \times 30 = 45 V$

And in $R C$ circuit voltage drop is given by

$V^{2} = V_{R}^{2} + V_{C}^{2}$

$V_{C} = \sqrt{V^{2} - V_{R}^{2}} = \sqrt{(120)^{2} - (45)^{2}} = 111 V$