In the circuit below the battery voltage is 12 V and the res

In the circuit below, the battery voltage is 12 V and the resistors have the following values: R1 is 8.0 Ohm, R2 is 4.0 Ohm, R3 is 21.0 Ohm, and R4 is 43.0 Ohm. With the switch open as shown... Find the current supplied by the battery. Find the voltage VM read by the meter. Find the power dissipated in R3. With the switch closed... Find the current supplied by the battery. Find the voltage VM read by the meter. Find the power dissipated in R3.

Solution

when the switch is open

R4 is not effective.

so

(b)the voltage read by the voltameter=i*R2 = 0.364x4= 1.456 volt

(c) power dissipated in R3= i² R3 = 0.364*0.364*21=2.78 watt

when the switch is closed

(a) the current supplied by the battery= 0.459amp

(R3 and R4 are in parrel so equivalent of R3 and R4 resistance is 14.11 ohm, total resistance of the circuit=8+4+14.11=26.11ohm) from v =i*R we have i= 0.459

(c) 0.459=i3+i2 ---(1) i2 is the current in R4

R4*i2=R3*i3

43*i2=21*i3

i2= 0.49i3

putting this value of i2 in eq (1)

0.459=1.49i3

i3=0.308amp

so power dissipated is i²R3 =0.308*0.308*21=1.99watt