Question

Two bulbs A & B rated (200V, 50W); (200V; 100W) are connected in series. A kid touches bulb A with its left hand and bulb B with its right hand. Which hand of the kid will be warmer?

• A. Left hand
• B. Right hand
• C. Both hands are equally warm
• D. The hands don’t get warm

Answer 1

The correct option is A Left hand

We know,
$P=VI$ where, P is the power, V is the voltage and I is the current.
We can also write, $P=\frac{V^2}{R}$
or $R=\frac{V^2}{P}$
For bulb A, $R_A=\frac{{200}^2}{50}=800\Omega$
For bulb B, $R_B=\frac{{200}^2}{100}=400\Omega$

P = VI
Replace (V = IR)
$\Rightarrow P=I^{2}R$
Since current is constant in series,
$\Rightarrow P\propto R$
More the resistance more the power and more the bulb glows.
Since, $R_A>R_B$, $P_A>P_B$
​Since the power consumed is greater by bulb A, the heat energy generated by the bulb A is higher. Hence, the left hand that touches the bulb A will be warmer.