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 $V _0$ is almost zero, then identify the correct statement.
(a) The emf of the battery is $1.5 V$ and its internal resistance is $1.5 \Omega$
(b) The value of the resistance $R$ is $1.5 \Omega$
(c) The potential difference across the battery is $1.5 V$ when it sends a current of $1000 mA$
(d) The emf of the battery is $1.5 V$ and the value of $R$ is $1.5 \Omega$
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Solution:
- Given circuit in a series combination of internal resistance of cell $(r)$ and external resistance $R$.
$\therefore$ Effective resistance in the circuit,
$$ R _{eff}=r+R $$
$\therefore$ Current in the circuit,
$$ I=\frac{E}{R+r} \text { or } E=I R+I r $$
Voltage difference across resistance $R$ is $V$, so
$$ E=V+I r $$
Now, from graph at $I=0, V=1.5 V$
From Eq. (i) at $I=0$,
$$ E=V=1.5 V $$
At $I=1000 mA$ (or $1 A$ ), $V=0$
From Eq. (i) at $I=1 A$ and $V=0$
$$ \Rightarrow \quad E=I \times r=r $$
From Eqs. (ii) and (iii), we can get
$$ \begin{array}{rlrl} & & r & =E=V=1.5 V \\ \therefore & r & =1.5 \Omega \end{array} $$
