Two resistances of same value, one capacitor and one micro ammeter is connected to a 5V battery of negligible internal resistance and a switch (s) as shown in the figure. When the switch is closed the micro ammeter shows a reading of \( 20\,\mu\mathrm{A} \). The value of R is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 250\,\mathrm{k}\Omega \)
In the figure the three branches (R with the microammeter, R with the capacitor, and the battery with the switch) are all joined between the same two junctions. So the branch containing the microammeter is in parallel with the battery and has the full 5 V across it, whatever happens in the capacitor branch. The battery has negligible internal resistance, and the microammeter is taken as ideal (negligible resistance).
By Ohm's law for the microammeter branch:
\( \displaystyle I = \frac{V}{R} \;\Rightarrow\; R = \frac{V}{I} \)
\( I = 20\,\mu\mathrm{A} = 20\times10^{-6}\,\mathrm{A} \)
\( \displaystyle R = \frac{5\,\mathrm{V}}{20\times10^{-6}\,\mathrm{A}} = 2.5\times10^{5}\,\Omega = 250\,\mathrm{k}\Omega \)
The capacitor branch only draws a brief charging current; in the steady state the capacitor blocks dc and that branch carries no current, so the reading stays \( 20\,\mu\mathrm{A} \).