Question
A capacitor is charged using an external battery with a resistance $$x$$ in series. The dashed line shows the variation of In $$I$$ with respect to time. If the resistance is changed to $$2x,$$ the new graph will be
A capacitor is charged using an external battery with a resistance $$x$$ in series. The dashed line shows the variation of In $$I$$ with respect to time. If the resistance is changed to $$2x,$$ the new graph will be
A.
$$P$$
B.
$$Q$$
C.
$$R$$
D.
$$S$$
Answer :
$$Q$$
Solution :
KEY CONCEPT : The current in $$RC$$ circuit is given by $$I = {I_0}{e^{ - \frac{t}{{RC}}}}$$
$$\eqalign{ & {\text{or}}\,lnI = ln{I_0} - \frac{t}{{RC}}{\text{ or }}lnI = \left( {\frac{{ - t}}{{RC}}} \right) + ln{I_0} \cr & lnI = \left( {\frac{{ - t}}{{RC}}} \right) + ln\left( {\frac{{{E_0}}}{R}} \right) \cr} $$
On comparing with $${\text{ }}y = mx + C$$
$${\text{Intercept }} = ln\left( {\frac{{{E_0}}}{R}} \right){\text{ and slope }} = - \frac{1}{{RC}}$$
When $$R$$ is changed to $$2R$$ then slope increases and current becomes less. New graph is $$Q.$$
KEY CONCEPT : The current in $$RC$$ circuit is given by $$I = {I_0}{e^{ - \frac{t}{{RC}}}}$$
$$\eqalign{ & {\text{or}}\,lnI = ln{I_0} - \frac{t}{{RC}}{\text{ or }}lnI = \left( {\frac{{ - t}}{{RC}}} \right) + ln{I_0} \cr & lnI = \left( {\frac{{ - t}}{{RC}}} \right) + ln\left( {\frac{{{E_0}}}{R}} \right) \cr} $$
On comparing with $${\text{ }}y = mx + C$$
$${\text{Intercept }} = ln\left( {\frac{{{E_0}}}{R}} \right){\text{ and slope }} = - \frac{1}{{RC}}$$
When $$R$$ is changed to $$2R$$ then slope increases and current becomes less. New graph is $$Q.$$
