A cell of e-m-f- 1-8 V and internal resistance 2 - is connected in series with an ammeter of resistance 0-7 - and a resistor of 4-5 - as shown in the figure- a What would be the reading of the ammeter- b What is the potential difference across the terminals of the cell-

(a) Here, E = 1.8 V r = 2 Ω R = 4.5 + 0.7 = 5.2 Ω Now, current, I = E/r+R = 1.8/2+5.2 = 0.25 A (b) Potential difference across the cell, V=IR = 0.25 × 5.2 = ...

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Standard XII

Physics

Simple Circuit

A cell of e.m...

Question

A cell of e.m.f. 1.8 V and internal resistance $2 Ω$ is connected in series with an ammeter of resistance $0.7 Ω$ and a resistor of $4.5 Ω$ as shown in the figure.
(a) What would be the reading of the ammeter?
(b) What is the potential difference across the terminals of the cell?

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Solution

(a) Here, $E = 1.8 V r = 2 Ω R = 4.5 + 0.7 = 5.2 Ω$

Now, current, $I = \frac{E}{r + R} = \frac{1.8}{2 + 5.2} = 0.25 A$

(b) Potential difference across the cell, $V = I R = 0.25 \times 5.2 = 1.3 V$

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A cell of e.m.f. 1.8 V and internal resistance 2Ω is connected in series with an ammeter of resistance 0.7Ω and a resistor of 4.5Ω as shown in the figure. (a) What would be the reading of the ammeter? (b) What is the potential difference across the terminals of the cell?

(a) Here, E=1.8 Vr=2 ΩR=4.5+0.7=5.2 Ω Now, current, I=Er+R=1.82+5.2=0.25 A (b) Potential difference across the cell, V=IR=0.25×5.2=1.3 V

A cell of e.m.f. 1.8 V and internal resistance $2 Ω$ is connected in series with an ammeter of resistance $0.7 Ω$ and a resistor of $4.5 Ω$ as shown in the figure.
(a) What would be the reading of the ammeter?
(b) What is the potential difference across the terminals of the cell?

(a) Here, $E = 1.8 V r = 2 Ω R = 4.5 + 0.7 = 5.2 Ω$

Now, current, $I = \frac{E}{r + R} = \frac{1.8}{2 + 5.2} = 0.25 A$

(b) Potential difference across the cell, $V = I R = 0.25 \times 5.2 = 1.3 V$