The supply voltage to room is 120 V. The resistance of the lead wires is `6Omega`. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

Options

A. no change
B. `10 V`
C. `20 V`
D. more than `10 V`

Answer

D

Explanation

`R_(60) = ( 120 xx 120 )/( 60) Omega = 240 Omega `
Current ` = (120)/( 240 + 6) A = (120)/( 246) A`
Voltage across bulb ` = (120)/(246) xx 240 V`
` = 117.1 V`
`R_(240) = ( 120 xx 120 )/( 240) Omega = 60 Omega`
Resistance of parallel combination is
`( 60 xx 240)/( 60 + 240) Omega = 48 Omega`
Total resistance is ` ( 48 + 6 ) Omega = 54 Omega`.
Current `I` is `120//54 A`. Voltage across parallel combination is
`(120)/(54) xx 48 V = 106.7 V`
Change in voltage is `(117.1 - 106.7)V = 10.4 V`.