MHT CET · Physics · Experimental Physics
In balanced meter bridge, the resistance of bridge wire is \(0.1 \Omega cm\). Unknown resistance \(X\) is connected in left gap and \(6 \Omega\) in right gap, null point divides the wire in the ratio \(2: 3\). Find the current drawn from the battery of 5 V having negligible resistance.
- A 1 A
- B 1.5 A
- C 2 A
- D 5 A
Answer & Solution
Correct Answer
(A) 1 A
Step-by-step Solution
Detailed explanation
\(\frac{l_1}{l_2}=\frac{2}{3}=\frac{v_1}{v_2}\)
Now \(v_1+v_2=v=5 V=I R\)
For potentiometer wire \(\frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{2}{3}\)
\(\frac{x}{6}=\frac{2}{3} \Rightarrow \frac{x}{\ell}=4 \frac{\Omega}{m}\)
The resistance of wire is \(\frac{0.1 \Omega}{cm}=\frac{1 \Omega}{m}\)
\(\therefore (x+1)=R_{ eff }=5\)
\(\therefore i=\frac{V}{R_{ eff }}=\frac{5}{5}=1 A\)
Now \(v_1+v_2=v=5 V=I R\)
For potentiometer wire \(\frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{2}{3}\)
\(\frac{x}{6}=\frac{2}{3} \Rightarrow \frac{x}{\ell}=4 \frac{\Omega}{m}\)
The resistance of wire is \(\frac{0.1 \Omega}{cm}=\frac{1 \Omega}{m}\)
\(\therefore (x+1)=R_{ eff }=5\)
\(\therefore i=\frac{V}{R_{ eff }}=\frac{5}{5}=1 A\)
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