Question

A resistance R and a capacitor C are connected in series to a source
V=Vo sin wt.
Find:
(a) The peak value of the voltage across the
(i) resistance and
(ii) capacitor.
b) The phase difference between the applied voltage and current.
Which of them is ahead ?​

Answer 1

Given that,

Source voltage $V=V_{0}\sin\omega t$

Resistance = R

Capacitor = C

(a). We need to calculate the peak value of the voltage across the resistance

Using formula of peak voltage across the resistance

$V_{p}= \dfrac{V_{rms}\times\sqrt{2}}{R}$

Put the value into the formula

$V_{p}=\dfrac{V_{0}\sin\omega t\sqrt{2}}{R}$

(II). We need to calculate the peak value of the voltage across the capacitor

Using formula of peak voltage across the capacitor

$V_{p}=\dfrac{V_{rms}\sqrt{2}}{C}$

Put the value into the formula

$V_{p}=\dfrac{V_{0}\sin\omega t\sqrt{2}}{C}$

(b). We need to find the phase difference between the applied voltage and current

We know that,

The voltage across the resistance then the voltage will be in phase with the current.

The voltage across the capacitance then the voltage must leg the current by 90°

So, The voltage across the resistance will be ahead.

Hence, (a). The peak value of the voltage across the resistance and capacitance are $\dfrac{V_{0}\sin\omega t\sqrt{2}}{R}$ and $\dfrac{V_{0}\sin\omega t\sqrt{2}}{C}$

(b). The voltage across the resistance will be ahead.

Answer 2

Given that,

Source voltage $V=V_{0}\sin\omega t$

Resistance = R

Capacitor = C

(a). We need to calculate the peak value of the voltage across the resistance

Using formula of peak voltage across the resistance

$V_{p}= \dfrac{V_{rms}\times\sqrt{2}}{R}$

Put the value into the formula

$V_{p}=\dfrac{V_{0}\sin\omega t\sqrt{2}}{R}$

(II). We need to calculate the peak value of the voltage across the capacitor

Using formula of peak voltage across the capacitor

$V_{p}=\dfrac{V_{rms}\sqrt{2}}{C}$

Put the value into the formula

$V_{p}=\dfrac{V_{0}\sin\omega t\sqrt{2}}{C}$

(b). We need to find the phase difference between the applied voltage and current

We know that,

The voltage across the resistance then the voltage will be in phase with the current.

The voltage across the capacitance then the voltage must leg the current by 90°

So, The voltage across the resistance will be ahead.

Hence, (a). The peak value of the voltage across the resistance and capacitance are $\dfrac{V_{0}\sin\omega t\sqrt{2}}{R}$ and $\dfrac{V_{0}\sin\omega t\sqrt{2}}{C}$

(b). The voltage across the resistance will be ahead.

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