No links please Q A galvanometer of resistance 45 ohm, shunted with a resistance of 5 ohm - Physics - Alternating Current

Question

No links please

Q A galvanometer of resistance 45 ohm, shunted with a resistance of
5 ohm is joined in series with a resistance of 84 ohm and a battery
of e m f 4 5 volt having internal resistance 1 5ohm Find the
current passing through (a) the battery, (b) the galvanometer, and
(c) the shunt

Sanjay Upadhyay · Accepted Answer

Dear student

Potential difference  through R1 in a circuit where R1, R2, R3 are in series is given by VR1=V×R1R1+R2+R3a) PD across the battery= 4
5×1 54 5+5+84+1 5=0
071VNow current through battery I=VR⇒I=0
0711 5=0
0473Ab) PD across the galvanometer= 4
5×454 5+5+84+1 5=2
13VNow current through the galvenometer I=VR⇒I=2
1345=0 0473Ab) PD across the shunt= 4
5×54 5+5+84+1 5=0
216VNow current through shunt I=VR⇒I=0
2365=0
0473ASince all the components are in series hence current is same for all
Regards

No links please Q. A galvanometer of resistance 45 ohm, shunted with a resistance of 5 ohm is joined in series with a resistance of 84 ohm and a battery of e.m.f. 4.5 volt having internal resistance 1.5ohm. Find the current passing through (a) the battery, (b) the galvanometer, and (c) the shunt.

No links please

Q. A galvanometer of resistance 45 ohm, shunted with a resistance of 5 ohm is joined in series with a resistance of 84 ohm and a battery of e.m.f. 4.5 volt having internal resistance 1.5ohm. Find the current passing through (a) the battery, (b) the galvanometer, and (c) the shunt.