Current Electricity Ques 28
- 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
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 $ $\Omega$
(2019, Main, 12 April I)
Show Answer
Answer:
Correct Answer: 28.(a)
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 $ $\quad$ …….(i)
Now, from graph at $I=0, V=1.5 $ $V$
From Eq. (i) at $I=0$,
$ E=V=1.5 $ $V $ $\quad$ …….(ii)
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 $ $\quad$ …….(iii)
From Eqs. (ii) and (iii), we can get
$ r =E=V=1.5$ $ V $
$\therefore \quad r =1.5 $ $\Omega$
BETA
