Question 654: Current Electricity

Question

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at length $$l$$
1
cm and $$l$$
2
cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to

Accepted Answer

(i) When key between the terminals 1 and 2 is plugged in,
P.D. across R = IR = $$k l_1$$
$$ \Rightarrow $$ R =$$ k l_1$$ as I = 1A
(ii) When key between terminals 1 and 3 is plugged in,
P.D. across (X + R) = I(X + R) = $$k l_2$$
$$ \Rightarrow $$  X + R = $$k l_2$$
$$ 	herefore $$ X = $$k (l_2 Ã¢ÂÂ l_1)$$
$$ 	herefore $$ R = $$kl_1$$ and X = $$k (l_2 Ã¢ÂÂ l_1)$$

Answer

$$k({l_2} - {l_1})$$ and $$k{l_2}$$

Answer

$$k{l_1}$$ and $$k({l_2} - {l_1})$$

Answer

$$k\left( {{l_2} - {l_1}} \right)$$ and $$k{l_1}$$

Answer

$$k{l_1}$$ and $$k{l_2}$$

Current Electricity

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