KCET · Physics · Alternating Current
A resistor and a capacitor are connected in series with an AC source. If the potential drop across the capacitor is \(5 \mathrm{~V}\) and that across resistor is \(12 \mathrm{~V}\), then applied voltage is
- A \(13 \mathrm{~V}\)
- B \(17 \mathrm{~V}\)
- C \(5 \mathrm{~V}\)
- D \(12 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(13 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Let the applied voltage be \(V\) volt.

Here, \(\mathrm{V}_{\mathrm{R}}=12 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=5 \mathrm{~V}\)
\(\begin{aligned}\therefore \mathrm{V} &=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{C}}^{2}}=\sqrt{(12)^{2}+(5)^{2}} \\
&=\sqrt{144+25}=\sqrt{169}=13 \mathrm{~V}\end{aligned}\)

Here, \(\mathrm{V}_{\mathrm{R}}=12 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=5 \mathrm{~V}\)
\(\begin{aligned}\therefore \mathrm{V} &=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{C}}^{2}}=\sqrt{(12)^{2}+(5)^{2}} \\
&=\sqrt{144+25}=\sqrt{169}=13 \mathrm{~V}\end{aligned}\)
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