In the circuit below the ammeter and voltmeter have resistan

In the circuit below, the ammeter and voltmeter have resistances of 5 [?] and 1000 [?] respectively. Determine the current flowing through the ammeter and the voltage drop across the voltmeter. Calculate the power dissipated in R1 and R2. Calculate the power supplied by the battery.

Solution

Potential difference V = 6 volt

resistances r ,R1,R2 are in series.

So, resultant resistance R = r +R1+R2

                                       = 2 ohm + 50 ohm+50 ohm

                                       =102 ohm

Current through the circuit i = V/R

                                        = 6 /102

                                        = 58.82 x10 ^-3 A

Potential across resistor R2 is V2 = V[R2/(r+R1+R2)]

                                                  = 6[50/(2+50+50)]

                                                   = 2.941 volt

voltmeter reading = 2.941 volt

ammeter reading = 58.82 x10 ^-3 A

(b).Power dissipated in R1 is P1 = i² R1

                                               = (58.82 x10 ^-3) ² (50)

                                               = 0.1729 watt

Power dissipated in R2 is P2 = i² R2

                                               = (58.82 x10 ^-3) ² (50)

                                               = 0.1729 watt

(c). Power supplied by the battery P = V ²/ R

                                                     = 6 ² /102

                                                     = 0.3529 watt