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Expert-verified Found in: Page 862 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # A short circuit to the grounded metal case of an appliance occurs as shown in Figure $${\rm{23}}{\rm{.60}}$$. The person touching the case is wet and only has a $${\rm{3}}{\rm{.00 k\Omega }}$$ resistance to earth/ground. (a) What is the voltage on the case if $${\rm{5}}{\rm{.00 mA}}$$ flows through the person? (b) What is the current in the short circuit if the resistance of the earth/ground wire is $${\rm{0}}{\rm{.200 \Omega }}$$? (c) Will this trigger the $${\rm{20}}{\rm{.0 A}}$$ circuit breaker supplying the appliance? (a) The voltage on the case is $$V = 15\;\;{\rm{V}}$$, if $${\rm{5}}{\rm{.00 mA}}$$ current flows through the person.

(b) The current in the short circuit is $$I = 75\;{\rm{A}}$$, if the resistance is $${\rm{0}}{\rm{.200 \Omega }}$$.

(c) Yes, the current will trigger the $${\rm{20}}{\rm{.0\;A}}$$ circuit breaker.

See the step by step solution

## Step 1: Concept Introduction

When we place a closed coil in a continuously changing magnetic field, a current is induced in the coil such that it opposes the change in the magnetic field. This phenomenon is known as electromagnetic induction.

## Step 2: Information Provided

• Resistance of the person: $${R_p} = 3.0\;{\rm{k}}\Omega$$.
• Amount of current flowing through the person:

\begin{aligned} {I_p} &= 5.00 {\rm{mA}}\\ &= \left( {\frac{{5.00}}{{1000}}} \right)\;{\rm{A}}\\ &= 0.005 {\rm{A}}\end{aligned}.

• Resistance of the earth/ground wire: $${R_g} = 0.200 \Omega$$.
• Current rating of the circuit breaker: $${I_c} = 20.0\;{\rm{A}}$$.

## Step 3: Calculation for Voltage

(a)

The voltage on the case will be equal to the voltage drop for the person’s body. For the resistance of body $${R_p} = 3.0\;{\rm{k}}\Omega$$ and the current through the body $${I_p} = 0.005 {\rm{A}}$$voltage can be calculated as –

\begin{aligned} {V_p} &= {I_p}{R_p}\\ &= 0.005\;{\rm{A}} \times 3000\;\Omega \\ &= 15\;{\rm{V}}\end{aligned}

Therefore, the value of voltage is found to be $${V_p} = 15\;{\rm{V}}$$.

## Step 4: Calculation for Current

(b)

The current through the grounding can be calculated using the relation –

\begin{aligned} {I_g} &= \frac{V}{{{R_g}}}\\ &= \frac{{15\;{\rm{V}}}}{{0.2\;\Omega }}\\ &= 75\;{\rm{A}}\end{aligned}

Therefore, the value of current through the grounding is $$I = 75\;{\rm{A}}$$.

## Step 5: Triggering of the Circuit Breaker

(c)

As the current rating for the circuit breaker is $$20.0\;{\rm{A}}$$, a grounding current of $$75\;{\rm{A}}$$ will trigger the circuit breaker.

Therefore, the circuit breaker will be triggered. ### Want to see more solutions like these? 