Question: Q. 3. Two cells of emfs $E_{1} & E_{2}$ and internal resistances $r_{1} & r_{2}$ respectively are connected in parallel. Obtain expressions for the equivalent.
(i) resistance and
(ii) emf of the combination U [CBSE Comptt. 2018]
Sol. Obtaining Expression for the equivalent
(i) resistance
(ii) emf
$$ \begin{array}{rlrl} & & \frac{1}{r} & =\frac{1}{r_{1}}+\frac{1}{r_{2}} \ \therefore & & r & =\frac{r_{1} r_{2}}{r_{1}+r_{2}} \ I & =I_{1}+I_{2} \ V & =E_{1}-I_{1} r_{1} \text { and } V=E_{2}-I_{2} r_{2} \ \therefore \quad & & I & =\left(\frac{E_{1}-V}{r_{1}}\right)+\left(\frac{E_{2}-V}{r_{2}}\right) \ \text { also } & & V & =E_{e q}-I r_{e q} \ \text { (i) } & & r_{e q} & =\frac{r_{1} r_{2}}{r_{1}+r_{2}} \ \text { (ii) } & & E_{e q} & \left.=\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1}+r_{2}}\right)-I\left(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\right) \ & & & \text { [CBSE Marking Scheme 2018] } \end{array} $$
Q. 4. First a set of n equal resistors of $R$ each is connected in series to a battery of emf $E$ and internal resistance $R$. A current $I$ is observed to flow. Then the $n$ resistors are connected in parallel to the same battery. It is observed that the current becomes 10 times. What is $n$ ?
A [SQP I 2017]
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Solution:
Ans.
$$ \begin{aligned} & I=\frac{E}{(R+n R)} \ & 10 I=\frac{E}{R+\frac{R}{n}} \ & \frac{1+n}{\left(1+\frac{1}{n}\right)}=10 \Rightarrow \frac{1+n}{(n+1)} n=10 \ & n=10 \end{aligned} $$
[CBSE Marking Scheme 2017]
Detailed Answer :
It is observed that an equivalent resistance of series combination is in series with internal resistance of a battery. Equivalent resistance of parallel combination will also be in series with internal resistance of battery.
Hence in series combination of resistors, current $I$ will be :
$$ \begin{equation*} I=\frac{E}{(R+n R)} \tag{i} \end{equation*} $$
In case of parallel combination, current $10 I$ will be :
$$ \begin{equation*} \frac{E}{\left(R+\frac{R}{n}\right)}=10 I \tag{ii} \end{equation*} $$
From the equations (i) and (ii),
$$ \frac{1+n}{\left(1+\frac{1}{n}\right)}=10 $$
$$ \begin{array}{ll} \text { So, } & 10=\frac{1+n}{(n+1)} n \ \text { Hence, } & n=10 \end{array} $$
