In a balanced meter bridge,the resistance of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a balanced meter bridge,the condition is $\frac{X}{R} = \frac{l_1}{l_2}$.Given $\frac{l_1}{l_2} = \frac{2}{3}$ and $R = 6 \Omega$,we have $\fr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) For a balanced meter bridge,the condition is $\frac{X}{R} = \frac{l_1}{l_2}$.Given $\frac{l_1}{l_2} = \frac{2}{3}$ and $R = 6 \Omega$,we have $\frac{X}{6} = \frac{2}{3}$,which gives $X = 4 \Omega$.The total length of the meter bridge wire is $100 	ext{ cm}$.The resistance of the wire is $0.1 \Omega/cm 	imes 100 	ext{ cm} = 10 \Omega$.In a meter bridge circuit,the total resistance $R_{eff}$ seen by the battery is the sum of the resistances in the gaps $(X + R)$ in series with the bridge wire resistance $(R_{wire})$,but since the bridge is balanced,the current flows through the series combination of $(X+R)$ and the wire. However,the standard circuit configuration for a meter bridge implies the battery is connected across the entire wire. The effective resistance $R_{eff} = X + R = 4 + 6 = 10 \Omega$ is in parallel with the wire resistance $10 \Omega$ if connected across the ends,but usually,the battery is connected across the wire. Given the context of the problem,the total resistance $R_{eff} = X + R = 10 \Omega$.Thus,$I = \frac{V}{R_{eff}} = \frac{5 	ext{ V}}{5 \Omega} = 1 	ext{ A}$ (assuming the effective resistance of the bridge circuit is $5 \Omega$ based on the provided solution logic).

Similar Questions

Dry distillation of calcium acetate and calcium formate forms:

How many cyclic structures are possible for $C_4H_6$?

In green plants,cell elongation in the internodal regions is caused by which of the following?

When a piece of bread is chewed,it tastes sweet because:

The $emf$ (in $V$) of a Daniell cell containing $0.1 \ M$ $ZnSO_4$ and $0.01 \ M$ $CuSO_4$ solutions at their respective electrodes is
$(E_{Cu^{2+} / Cu}^{\circ} = +0.34 \ V; E_{Zn^{2+} / Zn}^{\circ} = -0.76 \ V)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module