Question

A $220V$ and $800W$ electric kettle and three $220V$ and $100W$ bulbs are connected in parallel. On connecting this combination in parallel with $200V$ supply, the total current in the circuit will be:

• A. $0.15A$
• B. $5A$
• C. $5.5A$
• D. $4.55A$

Answer 1

The correct option is D $4.55A$

$P_1=\frac{V^2}{R_1}$

$P_2=\frac{V^2}{R_2}$

$⟹R_1=\frac{220\times 220}{800}$

$R_2=\frac{220\times 200}{100}$

$R_1=60.5\Omega$

$R_2=484\Omega$

$\frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{3}{R_2}$

$\frac{1}{R_{eff}}=\frac{2}{121}+\frac{3}{484}$

$=\frac{11}{484}$

$R_{eff}=44$

$I{R}_{eff}=200$

$I=\frac{200}{44}=4.55A$