A potentiometer wire of length \(1 \mathrm{~m}\) and resistance \(10 \Omega\) is connected in series with a cell of e.m.f \(2 \mathrm{~V}\) with internal resistance \(1 \Omega\) and a resistance box of a resistance \(R\) if potential difference between ends of the wire is \(1 \mathrm{~V}\) the value of \(R\) is. (A) \(4.5 \Omega\) (B) \(9 \Omega\) (C) \(15 \Omega\) (D) \(20 \Omega\)

Short Answer

Expert verified
The value of the resistance R in the circuit is \(9 \Omega\), which corresponds to option (B).

Step by step solution

01

Identify Known and Unknown Quantities

Let's list the given values: - Length of the potentiometer wire \(l = 1\) m - Resistance of the potentiometer wire \(r_p = 10\) ohms - Electromotive force (e.m.f) of the cell \(e = 2\) V - Internal resistance of the cell \(r_i = 1 \Omega\) - Potential difference between the ends of the wire \(V = 1\) V - Resistance box of resistance \(R\) The main task is to find the value of \(R\).
02

Use Ohm's Law and Kirchhoff's Laws in the Circuit

According to Ohm’s law, \[V = IR\] And, Kirchhoff’s voltage law states that the algebraic sum of the potential differences (voltages) in any closed loop or mesh in a network is always equal to zero. So, in our case: \[e - I(r_p + r_i + R) = 0\] We can replace \(V\) in Ohm's law with \(IR\) and substitute the given values: \[2 - I(10 + 1 + R) = 0\] We can write the potential difference across the potentiometer wire as: \[V = I r_p\] Substitute the given value: \[1 = I \cdot 10\] From this equation, we can find the current I: \[I = 0.1\mathrm{A}\]
03

Calculate the Value of R

Now, we can substitute the value of I in the Kirchhoff's equation: \[2 - 0.1(10 + 1 + R) = 0\] Solve for R: \[2 - (1.1 + 0.1R) = 0\] \[1.1 + 0.1R = 2\] \[0.1R = 1\] \[R = 10\] The total resistance in the circuit is \(R = 10\), and since we have a resistance box and internal resistance of cell, \[R_{total} = R + r_i\] \[R = R_{total} - r_i\] \[R = 10\Omega - 1\Omega\] \[R = 9\Omega\] The value of R is \(9 \Omega\), so the correct answer is option (B).

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