Current Electricity 2 Question 1

1. To verify Ohm’s law, a student connects the voltmeter across the battery as shown in the figure. The measured voltage is plotted as a function of the current and the following graph is obtained

(2019, Main, 12 April I)

If V0 is almost zero, then identify the correct statement.

(a) The emf of the battery is 1.5V and its internal resistance is 1.5Ω

(b) The value of the resistance R is 1.5Ω

(c) The potential difference across the battery is 1.5V when it sends a current of 1000mA

(d) The emf of the battery is 1.5V and the value of R is 1.5Ω

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Solution:

  1. Given circuit in a series combination of internal resistance of cell (r) and external resistance R.

Effective resistance in the circuit,

Reff=r+R

Current in the circuit,

I=ER+r or E=IR+Ir

Voltage difference across resistance R is V, so

E=V+Ir

Now, from graph at I=0,V=1.5V

From Eq. (i) at I=0,

E=V=1.5V

At I=1000mA (or 1A ), V=0

From Eq. (i) at I=1A and V=0

E=I×r=r

From Eqs. (ii) and (iii), we can get

r=E=V=1.5Vr=1.5Ω



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