A parallel combination of three resistors takes a current of 7.5 A from a 30 V battery. If the resistances of two resistors are 12 Ω and 10 Ω, then the resistance of the third resistor is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15 Ω
The equivalent resistance of the combination follows from Ohm's law, and for resistors in parallel the reciprocals add.
\( \displaystyle R_{\mathrm{eq}} = \frac{V}{I} = \frac{30\,\mathrm{V}}{7.5\,\mathrm{A}} = 4\,\Omega \)
\( \displaystyle \frac{1}{R_{3}} = \frac{1}{R_{\mathrm{eq}}} - \frac{1}{12} - \frac{1}{10} = \frac{1}{4} - \frac{1}{12} - \frac{1}{10} \)
\( \displaystyle \frac{1}{R_{3}} = \frac{15 - 5 - 6}{60} = \frac{4}{60} \;\Rightarrow\; R_{3} = 15\,\Omega \)
Shortcut: The currents add: \( 30/12 = 2.5\,\mathrm{A} \) and \( 30/10 = 3\,\mathrm{A} \), so the third resistor takes \( 7.5 - 5.5 = 2\,\mathrm{A} \) and \( R_3 = 30/2 = 15\,\Omega \).