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Chapter 12: Problem 1738

The potential difference between the terminals of a battery is $10 \mathrm{~V}\( and internal resistance \)1 \Omega\( drops to \)8 \mathrm{~V}$ when connected across an external resistor find the resistance of the external resistor. (A) \(40 \Omega\) (B) \(0.4 \Omega\) (C) \(4 \mathrm{M} \Omega\) (D) \(4 \Omega\)

Short Answer

Expert verified
The resistance of the external resistor is (D) \(4 \Omega\).

Step by step solution

01

Understand the given information and variables to use

We are given: - Potential difference between terminals of the battery, V1 = 10 V - Internal resistance of the battery, r = 1 Ω - Potential difference between terminals when connected to the external resistor, V2 = 8 V - The external resistor, R (find this)
02

Determine the current in the circuit when connected to the external resistor

When the battery is connected across the external resistor, the potential difference drops to 8 V. Using Ohm's law to calculate the current (I) in the circuit, we have: \(I = \frac{V_{2}}{R + r}\)
03

Determine the potential difference across the internal resistor

When the battery is connected across the external resistor, the potential difference across the internal resistor (r) is the difference between the potential differences in the open circuit and the closed circuit. Given this, we have: \(V_{r} = V_{1} - V_{2}\) where Vr is the potential difference across the internal resistor.
04

Calculate the potential difference across the internal resistor

Substituting the given values in the equation from Step 3, we get: \(V_{r} = 10 V - 8 V\) \(V_{r} = 2 V\)
05

Calculate the current in the circuit using the potential difference across the internal resistor

Using the potential difference across the internal resistor (Vr) and its resistance (r), we can calculate the current (I) in the circuit using Ohm's law: \(I = \frac{V_{r}}{r}\) Substituting the values, we get: \(I = \frac{2 V}{1 Ω}\) \(I = 2 A \)
06

Calculate the resistance of external resistor using Ohm's law

We already have calculated the current in the circuit and potential difference across the external resistor when connected. We will use Ohm's law to calculate the resistance of the external resistor: \( R = \frac{V_{2}}{I}\) Substituting the values, we get: \( R = \frac{8 V}{2 A}\) \(R = 4 Ω\) Since 4 Ω matches option (D), the correct answer is: (D) \(4 \Omega\)

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Most popular questions from this chapter

A heater boils \(1 \mathrm{~kg}\) of water in time \(\mathrm{t}_{1}\) and another heater boils the same water in time \(\mathrm{t}_{2}\) If both are connected in series, the combination will boil the same water in time. (A) $\left\\{\left(\mathrm{t}_{1} \mathrm{t}_{2}\right) /\left(\mathrm{t}_{2}+\mathrm{t}_{1}\right)\right\\}$ (B) $\left\\{\left(\mathrm{t}_{1} \mathrm{t}_{2}\right) /\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)\right\\}$ (C) \(\mathrm{t}_{1}+\mathrm{t}_{2}\) (D) \(2\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)\)

Which is the dimensional formula for conductance from the give below? (A) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{2}\) (B) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (D) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\)

The effective resistance of a n number of resistors connected in parallel in \(\mathrm{x} \mathrm{ohm}\). When one of the resistors is removed, the effective resistance becomes y ohm. The resistance of the resistor that is removed is.... (A) \(\\{(\mathrm{xy}) /(\mathrm{x}+\mathrm{y})\\}\) (B) \(\\{(\mathrm{xy}) /(\mathrm{y}-\mathrm{x})\\}\) (C) \((\mathrm{y}-\mathrm{x})\) (D) \(\sqrt{x y}\)

The resistance of the wire made of silver at \(27^{\circ} \mathrm{C}\) temperature is equal to \(2.1 \Omega\) while at \(100^{\circ} \mathrm{C}\) it is \(2.7 \Omega\) calculate the temperature coefficient of the resistivity of silver. Take the reference temperature equal to \(20^{\circ} \mathrm{C}\) (A) \(4.02 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (B) \(0.402 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (C) \(40.2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\) (D) \(4.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)

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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