Question

The emf of a battery is 30v. This battery is placed in a circuit with three resistors 430ohms, 680ohms, and 1100ohms where the 680ohm and 1100ohm resistors are in parallel and then connected in series to the 430ohm resistor and a switch. The terminal voltage of the battery is 7.4 volts.

a) Draw the circuit using appropriate symbols and labels.

b) How much current will flow in this circuit?

c) How much power would be dissipated by the 1100 ohm resistor?

Please show all calculations and explain!!

Answer 1

a)

For circuit please see : http://imgur.com/AYKr6Fm

b)

Here , as 680 Ohm and 1100 ohm are in Parallel ,

1/Rp = 1/680 + 1/1100

Rp = 420.2 Ohm

as Rp and 430 ohm are in series

Req = 420.2 + 430

Req= 850.2 Ohm

Using Ohm's law

V = I*R

I = 30/850.2

I = 0.0353 A

the current in the circuit is 0.0353 A

c)

Using current divider ,

I(1100) = 680 * 0.0353 /(1100+680)

I(1100) = 0.0135 A

Power = I^2 * R
Power dissipated in 1100 Ohm = 0.0135^2 * 1100

Power dissipated in 1100 Ohm =0.2 W

the Power dissipated in 1100 Ohm is 0.2 W