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Q54PE

Expert-verifiedFound in: Page 862

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.

**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.**

- 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: \({I_c} = 20.0\;{\rm{A}}\).

(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 \({V_p} = 15\;{\rm{V}}\).

(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 \(I = 75\;{\rm{A}}\).

(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.

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