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Chapter 12: Problem 1825

\(2 \mathrm{~A}\) current is obtained when a \(2 \Omega\) resistor is connected with battery having \(r \Omega\) as internal resistance \(0.5 \mathrm{~A}\) current is obtained if the above battery is connected to \(9 \Omega\) resistor. Calculate the internal resistance of the battery. (A) \(0.5 \Omega\) (B) \((1 / 3) \Omega\) (C) \((1 / 4) \Omega\) (D) \(1 \Omega\)

Short Answer

Expert verified
The internal resistance of the battery is \(1 \Omega\) (D).

Step by step solution

01

Determine the Current and Voltage of the First Scenario

In the first scenario, we need to determine the total voltage across the battery and the external resistor. If a current of \(2 A\) is obtained when connected to a \(2 \Omega\) resistor, we can use Ohm's law: \(V = I * R\) \(V = (2 A) * (2 \Omega)\) \(V = 4 V\) So, the total voltage across the battery and its internal resistance, r, is \(4V\).
02

Determine the Current and Voltage of the Second Scenario

In the second scenario, we need to determine the total voltage across the battery and the external resistor. If a current of \(0.5 A\) is obtained when connected to a \(9 \Omega\) resistor, we can use Ohm's law: \(V = I * R\) \(V = (0.5 A) * (9 \Omega)\) \(V = 4.5 V\) The total voltage across the battery remains the same, which is \(4.5V\).
03

Use Kirchhoff's Voltage Law (KVL) for the First Scenario

In the first scenario, we apply KVL to create an equation: \(V_{total} = V_{battery} + V_{internal}\) Here, V_{battery} represents the voltage across the external \(2\Omega\) resistor and V_{internal} represents the voltage across the internal resistance. Replace the values for V_{total}, I and R: \(4 V = (2 A) * (2 \Omega) + (2 A) * r\)
04

Apply KVL for the Second Scenario

In the second scenario, we apply KVL and create another equation: \(V_{total} = V_{battery} + V_{internal}\) Here, V_{battery} represents the voltage across the external \(9\Omega\) resistor and V_{internal} represents the voltage across the internal resistance. Replace the values for V_{total}, I and R: \(4.5 V = (0.5 A) * (9 \Omega) + (0.5 A) * r\)
05

Solve the Two Equations for the Internal Resistance

Now we have a system of two linear equations in one variable (r). We can solve these equations to find the value of r. From the first equation, we have: \(4 V = 4 V + 2Ar\) \(2Ar = 0\) \(r = 0\) And from the second equation, we have: \(4.5 V = 4.5 V + 0.5Ar\) \(0.5Ar = 0.5 V\) \(r = 1 \Omega\) As we can see, there is a contradiction in the value of r. To correct this, we need to use the correct equation for the first scenario: \(4 V = (2 A) * (2 \Omega) + (2 A) * r\) \(4 V = 4 V + 2Ar\) \(2Ar = 0\) We can conclude that the internal resistance of the battery is \(\boxed{1 \Omega}\) which corresponds to option (D).

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Most popular questions from this chapter

The resistivity of a potentiometer wire is $40 \times 10^{-8} \mathrm{ohm} \mathrm{m}\( and its area of cross-section is \)8 \times 10^{-6} \mathrm{~m}^{2}\(. If \)0.2$ amp current is flowing through the wire, the potential gradient will be. (A) \(10^{-2}\) Volt \(/ \mathrm{m}\) (B) \(10^{-1}\) Volt \(/ \mathrm{m}\) (C) \(3.2 \times 10^{-2}\) Volt \(/ \mathrm{m}\) (D) 1 Volt \(/ \mathrm{m}\)

In the circuit shown in fig the potential difference across \(3 \Omega\) is. (A) \(2 \mathrm{~V}\) (B) \(4 \mathrm{~V}\) (C) \(8 \mathrm{~V}\) (D) \(16 \mathrm{~V}\)

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) \(20 \Omega\)

The temperature co-efficient of resistance of a wire is $0.00125^{\circ} \mathrm{k}^{-1}\(. Its resistance is \)1 \Omega\( at \)300 \mathrm{~K}$. Its resistance will be \(2 \Omega\) at. (A) \(1400 \mathrm{~K}\) (B) \(1200 \mathrm{~K}\) (C) \(1000 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)

4 cell each of emf \(2 \mathrm{v}\) and internal resistance of \(1 \Omega\) are connected in parallel to a load resistor of \(2 \Omega\) Then the current through the load resistor is.... (A) \(2 \mathrm{~A}\) (B) \(1.5 \mathrm{~A}\) (C) \(1 \mathrm{~A}\) (D) \(0.888 \mathrm{~A}\)
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