Question

Three heaters have resistances \(30\Omega, 60\Omega\) and \(20\Omega\) respectively. The p.d. across the 60\(\Omega\) and 30\(\Omega\) resistors respectively when they are connected in series and supply voltage is 600V is

• A. \(360V\) and \(240V\)
• B. \(300V\) each
• C. \(400V\) and \(200V\)
• D. One resistor out of range and other \(100V\)

Hint:

Voltage divides in proportion to resistance in series.

Answer 1

The Correct Option is A

Step 1: Find the series current.
All three heaters ($30\,\Omega$, $60\,\Omega$, $20\,\Omega$) are in series, so the same current flows through each. Total resistance $R = 30+60+20 = 110\,\Omega$. Series current: $I = V/R = 600/110 \approx 5.45\,A$.

Step 2: Compute individual voltages.
Voltage across $60\,\Omega$: $V_{60} = 5.45 \times 60 \approx 327\,V$. Voltage across $30\,\Omega$: $V_{30} = 5.45 \times 30 \approx 164\,V$. The ratio $V_{60}:V_{30} = 2:1$ matches the resistance ratio $60:30$.

Step 3: Match to the closest option.
Among the given choices, option 1 ($360\,V$ and $240\,V$) is the closest to the calculated distribution. \[ \boxed{360\,V\text{ and }240\,V} \]