Question 75: Current Electricity

Question

To the potentiometer wire of length L and 10$$\Omega$$ resistance, a battery of emf 2.5 V and a resistance R are connected in series. If a potential difference of 1 V is balanced across L/2 length, the value of R in $$\Omega$$ will be

Accepted Answer

The given situation is shown below.
Resistance of potentiometer wire, $$R_P=10~\Omega$$
Emf, $$E=2.5~V$$
Total resistance, $$R'=R_P+R=(10+R)~\Omega$$
Current thruogh potentiometer wire,
$$I=\frac{E}{R_P}=\frac{2.5}{10+R}$$
Since, potential difference across length $$AL=\frac{L}{2}$$,
$$V = I\left( {{{{R_P}} \over 2}} ight)$$
$$ \Rightarrow 1 = {{2.5} \over {10 + R}} 	imes {{10} \over 2}$$
$$ \Rightarrow 10 + R = 12.5$$
$$ \Rightarrow R = 2.5~\Omega $$

Answer

1

Answer

1.5

Answer

2

Answer

2.5

Current Electricity

To the potentiometer wire of length L and 10$\Omega resistance, a battery of emf 2.5 V and a resistance R are connected in series. If a potential difference of 1 V is balanced across L/2 length, the value of R in \Omega$ will be

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