Three resistors of resistances 2r, 3r and 5r are connected in series across a 20 V source. If the smallest resistance is \( 4\,\Omega \), what is the current in the circuit?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1 A
In a series combination the same current flows through every resistor, and the voltages across them add. So the equivalent resistance is the sum of the individual resistances. The current is the source voltage divided by this sum.
The smallest resistance is \( 2r \), so:
\( 2r=4\,\Omega \)
\( \Rightarrow\; r=2\,\Omega \)
The equivalent resistance is:
\( \begin{aligned} R_{\mathrm{eq}} &= 2r+3r+5r \\ &= 10\times2\,\Omega \\ &= 20\,\Omega \end{aligned} \)
The current in the circuit is:
\( \begin{aligned} I &= \frac{V}{R_{\mathrm{eq}}} \\ &= \frac{20\,\mathrm{V}}{20\,\Omega} \\ &= 1\,\mathrm{A} \end{aligned} \)