/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 A capacitor connected to a \(230... [FREE SOLUTION] | 91Ó°ÊÓ

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A capacitor connected to a \(230 \mathrm{~V}, 50 \mathrm{~Hz}\) supply draws a 15 A current. What current will it draw when the capacitance and frequency are both reduced to half?

Short Answer

Expert verified
The new current is calculated using Ohm's Law and the newly calculated reactance, resulting in the current being the same as the original.

Step by step solution

01

Calculation of Original Reactance

Use the formula for reactance \(X_C = V/I\) where V is the voltage and I is the current. This gives \(X_C = 230/15\).
02

Calculation of Original Capacitance

Use the formula for capacitance \(C = 1/(2\(\pi\)fX_C)\) where f is the frequency and \(X_C\) is the reactance. This gives \(C = 1/(2\(\pi\)50\(\times\)X_C)\).
03

Calculation of New Reactance

Use the formula for reactance \(X_C = 1/(2\(\pi\)fC)\) but this time halve the frequency and the capacitance as the problem states. This gives \(X_C = 1/(2\(\pi\)25\(\times\)(C/2))\).
04

Calculation of New Current

Finally, use Ohm's Law to find the new current: \(I = V/X_C\) where V is the voltage and \(X_C\) is the new reactance.

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Key Concepts

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

Reactance of a Capacitor
In an alternating current (AC) circuit, a capacitor does not directly resist the flow of current in the same way a resistor would. Instead, it provides reactance, a form of opposition that capacitors exhibit to the change in voltage. Unlike resistance, reactance can vary with the frequency of the electrical signal.
Reactance is crucial in understanding how capacitors behave in AC circuits. The formula for the reactance of a capacitor, often denoted as \(X_C\), is \(X_C = \frac{1}{2\pi fC}\), where \(f\) is the frequency of the AC supply and \(C\) is the capacitance in farads (F). A key feature of capacitive reactance is that it decreases with an increase in either frequency or capacitance. This means that a capacitor will allow more current to pass through as the frequency rises or as its capacitance becomes larger.
Capacitance Formula
Capacitance, the ability of a capacitor to store charge, is a central concept in understanding the behavior of capacitors in circuits. The capacitance of a capacitor can be calculated using the formula \(C = \frac{Q}{V}\), where \(Q\) is the electric charge stored on the capacitor, measured in coulombs (C), and \(V\) is the voltage across the capacitor, measured in volts (V).
However, when working with AC circuits, it is often more useful to express capacitance in terms of reactance and frequency. The rearranged formula from the reactance definition, \(C = \frac{1}{2\pi fX_C}\), directly relates capacitance to the AC frequency and the capacitive reactance. The unit of capacitance is the farad, which is a large unit, so capacitors are usually sized in microfarads (\(\mu F\)), nanofarads (nF), or picofarads (pF).
Ohm's Law
Ohm's Law is a fundamental principle in the field of electronics and electrical engineering, providing the relationship between voltage, current, and resistance in an electrical circuit. The law is usually stated as \(I = \frac{V}{R}\), where \(I\) is the current flowing through the conductor (in amperes), \(V\) is the voltage measured across the conductor (in volts), and \(R\) is the resistance of the conductor (in ohms, \(\Omega\)).
In AC circuits, Ohm's Law is also applied using reactance instead of resistance for components like capacitors and inductors. For capacitors, the law is adapted to \(I = \frac{V}{X_C}\), where \(X_C\) represents the capacitive reactance. This formula is crucial for determining how much current will flow through a capacitor at a given voltage and frequency. Understanding Ohm's Law allows students to analyze not only simple direct current (DC) circuits but also more complex AC circuits involving capacitive and inductive elements.

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Most popular questions from this chapter

A capacitive load takes \(10 \mathrm{kVA}\) and \(5 \mathrm{kVAR}\), when connected to a \(200 \mathrm{~V}, 50 \mathrm{~Hz}\) ac supply. Calculate (i) resistance, (ii) capacitance, (iii) active power, and (iv) pf.

Two coils \(A\) and \(B\) are connected in series across a \(200 \mathrm{~V}, 50 \mathrm{~Hz}\) ac supply. The power input to the circuit is \(2 \mathrm{~kW}\) and \(1.15 \mathrm{kVAR}\). If the resistance and reactance of the coil \(A\) are \(5 \Omega\) and \(8 \Omega\) respectively, calculate resistance and reactance of the coil \(B\). Also, calculate the active power consumed by coils \(A\) and \(B\).

An emf of \(100 \mathrm{~V}, 50 \mathrm{~Hz}\) is applied to an impedance \(Z_{1}=8+j 6 \Omega\). An impedance \(Z_{2}\) is added in series with \(Z_{1}\). The current becomes half of the original and leads it by \(20^{\circ}\). Determine \(Z_{2}\).

A resistor and capacitor are connected in series with a variable inductor. When the circuit is connected to a \(220 \mathrm{~V}, 50 \mathrm{~Hz}\) supply, the maximum current obtained by varying the inductance is \(0.314 \mathrm{~A}\). The voltage across the capacitor is \(800 \mathrm{~V}\). Calculate the resistance, inductance and capacitance of the circuit.

A voltage of \(200 \angle 53.13^{\circ} \mathrm{V}\) is applied across two impedances in parallel. The values of the impedances are \((12+j 16) \Omega\) and \((10-j 20) \Omega\). Determine kVA, kVAR and \(\mathrm{kW}\) in each branch and the \(\mathrm{pf}\) of the whole circuit.

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