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Chapter 12: Problem 15

Two lamps, one rated \(100 \mathrm{~W}\) at \(220 \mathrm{~V}\), and the other \(60 \mathrm{~W}\) at \(220 \mathrm{~V}\), are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is \(220 \mathrm{~V} ?\)

Short Answer

Expert verified
The total current drawn from the line is 0.727 A.

Step by step solution

01

Determine the resistance of the 100 W lamp

Use the formula for power: \[ P = \frac{V^2}{R} \] Rearrange this to find the resistance \( R \): \[ R = \frac{V^2}{P} \] Substitute the values for the 100 W lamp: \( V = 220 \mathrm{~V} \) and \( P = 100 \mathrm{~W} \) \[ R_{100W} = \frac{220^2}{100} = 484 \Omega \]
02

Determine the resistance of the 60 W lamp

Use the same formula: \[ R = \frac{V^2}{P} \] Substitute the values for the 60 W lamp: \( V = 220 \mathrm{~V} \) and \( P = 60 \mathrm{~W} \) \[ R_{60W} = \frac{220^2}{60} = 806.67 \Omega \]
03

Calculate the current through the 100 W lamp

Use Ohm's law: \[ I = \frac{V}{R} \] For the 100 W lamp, \( R = 484 \Omega \) and \( V = 220 \mathrm{~V} \): \[ I_{100W} = \frac{220}{484} = 0.454 \mathrm{~A} \]
04

Calculate the current through the 60 W lamp

Again, use Ohm's law: \[ I = \frac{V}{R} \] For the 60 W lamp, \( R = 806.67 \Omega \) and \( V = 220 \mathrm{~V} \): \[ I_{60W} = \frac{220}{806.67} = 0.273 \mathrm{~A} \]
05

Determine the total current drawn from the line

In a parallel circuit, the total current is the sum of the currents through each component: \[ I_{total} = I_{100W} + I_{60W} \] Substitute the calculated currents: \[ I_{total} = 0.454 \mathrm{~A} + 0.273 \mathrm{~A} = 0.727 \mathrm{~A} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental principle in electrical engineering. It describes the relationship between voltage (\text{V}), current (\text{I}), and resistance (\text{R}) in an electrical circuit. The formula is expressed as:

\( V = I \cdot R \).

This means that the voltage across a conductor is directly proportional to the current through it, provided the temperature remains constant. To find the current, you rearrange the formula to

\( I = \frac{V}{R} \).

In this exercise, Ohm's Law helps us calculate the current through each lamp based on its resistance and the supplied voltage.
Power Formula
The power formula relates power (\text{P}), voltage (\text{V}), and resistance (\text{R}) in a circuit. The power consumed by an electrical device is given by:

\( P = \frac{V^2}{R} \).

This formula can be rearranged to find the resistance when power and voltage are known:

\( R = \frac{V^2}{P} \).

Using this rearranged formula, we calculated the resistance of both the 100 W and 60 W lamps in the exercise. Knowing the resistance then allowed us to find the current using Ohm's Law.
Parallel Circuit
A parallel circuit is one where multiple components are connected across the same voltage source. Each component in a parallel circuit has the same voltage across it.

The total current in a parallel circuit is the sum of the currents through each component. This is important because the current in each branch can be different, but they all add up to form the total current drawn from the source.

In the exercise, the two lamps are connected in parallel to a 220 V supply. By finding the individual currents through each lamp and then summing them, we determined the total current drawn from the mains.

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Most popular questions from this chapter

Compare the power used in the \(2 \Omega\) resistor in each of the following circuits: (i) a \(6 \mathrm{~V}\) battery in series with \(1 \Omega\) and \(2 \Omega\) resistors, and (ii) a \(4 \mathrm{~V}\) battery in parallel with \(12 \Omega\) and \(2 \Omega\) resistors.

Show how you would connect three resistors, each of resistance \(6 \Omega\), so that the combination has a resistance of (i) \(9 \Omega\), (ii) \(4 \Omega\).

A hot plate of an electric oven connected to a \(220 \mathrm{~V}\) line has two resistance coils \(\mathrm{A}\) and \(\mathrm{B}\), each of \(24 \Omega\) resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

A battery of \(9 \mathrm{~V}\) is connected in series with resistors of \(0.2 \Omega, 0.3 \Omega, 0.4 \Omega, 0.5 \Omega\) and \(12 \Omega\), respectively. How much current would flow through the \(12 \Omega\) resistor?

A piece of wire of resistance \(R\) is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is \(R^{\prime}\), then the ratio \(R / R^{\prime}\) is - (a) \(1 / 25\) (b) \(1 / 5\) (c) 5 (d) 25
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