A circuit has a section $ A B $, shown in the figure. The emf of the source is equal to $ 10 \mathrm{~V} $ and the capacitance of capacitor are equal to $ C_{1}=1.0 \mu \mathrm{F} $ and $ C_{2}=2.0 \mu \mathrm{F} $. If the potential difference between points $ A $ and $ B, \phi_{\mathrm{A}}-\phi_{\mathrm{B}}=5.0 \mathrm{~V} $, then the voltages $ V_{1} $ and $ V_{2} $ across each capacitors, $ C_{1} $ and $ C_{2} $, respectively, are

Question

A circuit has a section $ A B $, shown in the figure. The emf of the source is equal to $ 10 \mathrm{~V} $ and the capacitance of capacitor are equal to $ C_{1}=1.0 \mu \mathrm{F} $ and $ C_{2}=2.0 \mu \mathrm{F} $. If the potential difference between points $ A $ and $ B, \phi_{\mathrm{A}}-\phi_{\mathrm{B}}=5.0 \mathrm{~V} $, then the voltages $ V_{1} $ and $ V_{2} $ across each capacitors, $ C_{1} $ and $ C_{2} $, respectively, are, $ V_{1}=\frac{5}{3} \mathrm{~V}, V_{2}=\frac{10}{3} \mathrm{~V} $, $ V_{1}=\frac{10}{3} \mathrm{~V}, V_{2}=\frac{10}{3} \mathrm{~V} $, $ V_{1}=\frac{10}{3} \mathrm{~V}, V_{2}=\frac{5}{3} \mathrm{~V} $, none of these

MARKS App · Accepted Answer

Let the potential at point A be V A and at point B be V B . From the question, we know that, V A - V B = 5 … i . Let us write the voltage in the circuit starting from point A . Voltage drop across capacitor is given by V = q C . Therefore, ⇒ V A + q C 1 - 10 + q C 2 = V B . Substituting the values and using equation i , - 5 = q 1 + q 2 - 10 ⇒ q = 10 3 μC Hence, the voltage drop across C 1 , V 1 = q C 1 = 10 3 × 1 = 10 3 V . And voltage drop across C 2 , V 2 = q C 2 = 10 3 × 2 = 5 3 V .

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