Question

A galvanometer together with an unknown resistance in series is connected across two identical batteries each of 1.5 volts when the batteries are connected in series the galvanometer records a current of 1 m and when batteries are connected in parallel the current is 16 and what is the internal resistance of each battery

Answer 1

Answer:

Explanation:

The net resistance = R+2r

Potential Difference(V) = 3V

As from the ohms law

3 = 1(R+2r)---------- 1.)

The net resistance in second case = R+r/2

V = 1.5V I =6A

1.5 = 6(R+r/2)

¼ =R+r/2 ---------(2.)

3r/2 = 11/4

r = 22/12

r =1.83 ohm

Answer 2

Answer:

The internal resistance is $\dfrac{1}{3}$

Explanation:

Given that,

Voltage = 1.5 volt

Current in series = 1 A

Current in parallel = 0.6 A

Let R be the combine resistance of galvanometer and an unknown resistance and r the internal resistance of each battery.

When the batteries, each of emf E are connected in series

The net emf  = 2E

The net internal resistance = 2r

We need to calculate the resistance and internal resistance

Using formula of current

$i_{1}=\dfrac{2E}{R+2r}$

Put the value into the formula

$1=\dfrac{2\times1.5}{R+2r}$

$R+2r=3.0$....(I)

When the batteries are connected in parallel, the emf remains E and net internal resistance become $\dfrac{r}{2}$

We need to calculate the current

Using formula of current

$i_{2}=\dfrac{E}{R+\dfrac{r}{2}}$

$i_{2}=\dfrac{2E}{2R+r}$

$2R+r=\dfrac{2E}{i_{2}}$

Put the value intro the formula

$2R+r=\dfrac{2\times1.5}{0.6}$

$2R+r=5$....(II)

From equation (I) and (II)

$r=\dfrac{1}{3}\ \Omega$

Hence, The internal resistance is $\dfrac{1}{3}$.