Question

A battery having an emf of 12V and an internal resistance of \[2\Omega \] is connected to
Another battery having an emf of 18 volt and an internal resistance of $2\Omega $ in such a way that they are opposing each other and the circuit is closed. Calculate the following
1. Current flowing in the circuit.
2. Electrical Power in the two batteries
3. Terminal voltage of the two batteries
4. Electrical power consumed in the batteries

Answer 1

Hint: The approach to solve this question is
1. Apply algebra in batteries and internal resistance separately to a single battery with single internal resistance
2. Remember ohm’s Law it is always helpful in solving questions related batteries and internal resistances by the expression ${i_{net}} = \dfrac{{{E_{net}}}}{{{r_{net}}}}$
3. Formula for terminal voltage is $V = E - IR$
4. For calculating power use $\dfrac{{{V^2}}}{R}$

In the solution part we will elaborate these terms

Complete step by step answer:
1. Let the given internal resistances of the two battery ${E_1}$ and ${E_2}$ be ${r_1}$ and ${r_2}$ According to the question the given values are :
${r_1}$= $2\Omega $
${r_2}$$ = 2\Omega $
${E_1}$$ = 12V$
${E_2} = 18V$
So,
Let us first discuss the algebra in Emf and internal resistance
${E_{net}} = {E_2} - {E_1}$
${r_{net}} = {r_1} + {r_2}$
 Where ${E_{net}}$ is the net emf we get by considering the sign convention of each battery.
 And ${r_{net}}$ is the net resistance throughout the circuit according to this question we have internal resistance in parallel hence We will apply submission formula for resistances in series
By using the above formula, we put the values of ${E_1}$ and ${E_2}$
$
  {E_{net}} = {E_2} - {E_1} = 18 - 12 = 6V \\
  {r_{net}} = {r_1} + {r_2} = 4\Omega \\
 $
Now using ohm's Law as mentioned above in Hint section
${i_{net}} = \dfrac{{{E_{net}}}}{{{r_{net}}}}$$ = \dfrac{6}{4} = 1.5A$
Hence the net current is 1.5 A
Formula for power is :
Power=P \[ = \dfrac{{{E^2}}}{r}\]
Where E is the emf of the battery and r is the internal resistance
Now we calculate power for the given Emf’s
For 12V battery \[P = \dfrac{{{{12}^2}}}{2} = 72W\]
For 18V battery $P = \dfrac{{{{18}^2}}}{2} = 162W$
1. Formula for Calculating terminal voltage is $V = E - iR$
If the current is entering from negative terminal then take $ + i$ and if entering from the positive terminal take $ - i$
For 12V battery terminal voltage $ = {V_{{T_1}}} = 12 + (1.5)(2) = 15V$
For 18V battery terminal voltage $ = {V_{{T_2}}} = 18 - (1.5)(2) = 15V$

2. As discussed above formula for power consumed is $\dfrac{{{V^2}}}{r}$ where V is the terminal voltage
Power consumed by 12V $ = \dfrac{{{{15}^2}}}{2} = 112.5W$
Power consumed by 18V $ = \dfrac{{{{15}^2}}}{2} = 112.5W$

Note: Clearly power consumed by both the Batteries are equal because whenever we connect two batteries of unequal emf to try make their power equal by transferring energy and hence we get the same power.