Solution:
The equivalent resistance in series is \(R_s = 5R\) and in parallel is \(R_p = \frac{R}{5}\). Since current \(I = \frac{V}{R_{eq}}\), we have \(\frac{I_p}{I_s} = \frac{R_s}{R_p} = \frac{5R}{R/5} = 25\). Thus, \(n = 25\).
Five resistors, each of resistance \(R\) are connected in series to an ideal battery. When these resistors are connected in parallel to the same battery, the current through the battery is increased \(n\) times. The value of \(n\) is
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