Explanation
E = V + Ir
V = 50V
E = 50+11r …………(i)
For V = 60V
E = 60 + r …………(ii)
On solving equation (i) and (ii)
E = 61 V
Applying KCL,
5 + 4 – 3 – 5 + I = 0
I = - 1A
V = E – ir = 10 – 0.5× 3 = 8.5 V
Net emf = E – E = 0
Current through circuit = 0
V = E + ir
= 2 + 5 × 0.1 = 2.5 V
The resistance is connected in parallel with the voltmeter. The Effective resistance decreases.
R = V/I.
With increase in resistance, current will decreases and vice versa.
Therefore both Ammeter and voltmeter reading increases.
Potentiometer works on null deflection method. In balance condition no current flows in secondary circuit
No current is drawn from the cell whose emf is to be measured in potentiometer whereas in voltmeter current is drawn from the cell. So, emf of a cell can be measured accurately with potentiometer.
A high resistance is connected in series with galvanometer to convert it into a voltmeter.
Due to high resistance of voltmeter, connected in series the effective resistance of circuit will increase and hence current in circuit will decrease. Due to which the ammeter and voltmeter will not be damaged.
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