Question

Six lead-acid types of secondary cells each of emf 2.0V and internal resistance 0.015 $\Omega $ are joined in series to provide a supply to a resistance of 8.5 $\Omega $. What are the current drawn from the supply and its terminal voltage?
A. 1.4A, 11.9V
B. 1.8A, 11.9V
C. 1.4A, 12.9V
D. 1.8A, 12.9V

Answer 1

Hint: Obtain the expression to find the total resistance of a system with more than one resistance connected in series. Find the total resistance and then find the total voltage of the system. From this data find the current through the system using the Ohm’s law. Then from the current find the terminal voltage.

Complete step-by-step answer:
Six lead acid cells are connected in series to a resistance.
The emf of each of the cells is 2.0V.
The internal resistances of the cells are 0.015 $\Omega $.
The resistance to which the cells supply power is 8.5 $\Omega $.
When two or more than two resistances are connected in series, the total resistance of the system can be found by simply adding the resistances to each other.
In the given system the six cells are connected in series which have their own internal resistance of 0.015 $\Omega $ and they are connected to a resistance of 8.5 $\Omega $. So, the total resistance of the system will be,
$\begin{align}
  & R=6\times 0.015+8.5 \\
 & R=8.59\Omega \\
\end{align}$
Each cell has an emf of 2 V. so, the total emf of the system will be,
$V=6\times 2V=12V$
So, the current in the circuit can be found out as,
$I=\dfrac{V}{R}=\dfrac{12V}{8.59\Omega }=1.4A$
So, the terminal voltage can be defined as
${{V}_{t}}=1.4A\times 8.5A=11.9V$

So, the correct answer is “Option A”.

Note: when two or more resistance are connected in series in a circuit, the total resistance of the system is given by adding all the resistances. When two or more resistances are connected in parallel to each other in a circuit, the total resistance of the system is given as,
$\dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+...$