What should be the value of resistance `R` in the circuit shown in figure so that the electric bulb consumes the rated power?

Answer

6

Explanation

Resistance of electric bulb, `R' = (V^(2))/(P) = (3^(2))/(0.5) = 18 Omega`
current through bulb, `l_(1) = (P)/(V) = (0.5)/(3) = (1)/(6) A`
current through resistance `R, I_(2) = (3)/(R )A`
Total current in the circuit `= l_(1) + l_(2)`
`= (1)/(6) + (3)/(R) = ((R + 18))/(6 R) A`
Total resistance of circuit
`= 0.5 + 4+ (18 xx R)/(18 + R) = 4.5 + (18 R)/(18 + R)`
`:.` Current in circuit `= (6)/(4.5 + (18 R)/(18+ R)) = (R+ 18)/(6 R)` [from (i)]
On solving we get `R = 6 Omega`