Question

(a) Find the voltage drop in an extension cord having a \({\bf{0}}{\bf{.0600}}\;{\bf{\Omega }}\) resistance and through which \({\bf{5}}{\bf{.00}}\;{\bf{A}}\) is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of \({\bf{0}}{\bf{.300}}\;{\bf{\Omega }}\). What is the voltage drop in it when \({\bf{5}}{\bf{.00}}\;{\bf{A}}\) flows? (c) Why is the voltage to whatever appliance is being used reduced by this amount? What is the effect on the appliance?

Short answer

1. The voltage drop in the extension cord is \(0.300\;{\rm{V}}\).
2. Thevoltage drop in the thinner wire is \(1.50\;{\rm{V}}\).
3. The voltage to whatever appliance is used is reduced by this amount because the current flow through the appliance would be safer. The appliances might not work efficiently on the applied voltage.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The electrical resistance of an extension cord is,\({R_1} = 0.0600\;{\rm{\Omega }}\).
• The electrical current flowing through the cord is,\({I_1} = 5.00\;{\rm{A}}\).
• The electrical resistance of a thinner wire is,\({R_2} = 0.300\;{\rm{\Omega }}\).
• The electrical current flowing through the thinner wire is, \({I_2} = {I_1} = 5.00\;{\rm{A}}\).

Significance of Ohm’s law and the electrical voltage drop

Ohm's law states that the voltage across a current-carrying resistor equals the product of resistance and the current. Connecting wires are conductors having a non-zero resistance; it means there will be a voltage across the wire when the current flows. This voltage will reduce the voltage across the appliance connected through the wires.

(a) Determination of the voltage drop in the extension cord

The expression to calculate the voltage drop in the extension cordis expressed as,

\({V_1} = {I_1}{R_1}\)

Here,\({V_1}\)is the voltage drop in the extension cord.

Substitute all the known values in the above equation.

\(\begin{align}{V_1} &= \left( {5.00\;{\rm{A}}} \right)\left( {0.0600\;{\rm{\Omega }}} \right)\\ &= 0.300\;{\rm{A}} \cdot {\rm{\Omega }}\\ &= \left( {0.300\;{\rm{A}} \cdot {\rm{\Omega }}} \right)\left( {\frac{{1\;{\rm{V}}}}{{1\;{\rm{A}} \cdot {\rm{\Omega }}}}} \right)\\ &= 0.300\;{\rm{V}}\end{align}\)

Thus, the voltage drop in the extension cord is \(0.300\;{\rm{V}}\).

(b) Determination of the voltage drop in the thinner wire

The expression to calculate the voltage drop in the thinner wire is expressed as,

\({V_2} = {I_2}{R_2}\)

Here,\({V_2}\)is the voltage drop in the extension cord.

Substitute all the known values in the above equation.

\(\begin{align}{V_2} &= \left( {5.00\;{\rm{A}}} \right)\left( {0.300\;{\rm{\Omega }}} \right)\\ &= 1.50\;{\rm{A}} \cdot {\rm{\Omega }}\\ &= \left( {1.50\;{\rm{A}} \cdot {\rm{\Omega }}} \right)\left( {\frac{{1\;{\rm{V}}}}{{1\;{\rm{A}} \cdot {\rm{\Omega }}}}} \right)\\ &= 1.50\;{\rm{V}}\end{align}\)

Thus, the voltage drop in the thinner wire is \(1.50\;{\rm{V}}\).

(c) Determination of the reason behind the reduced voltage to the appliance and the effect of the reduced voltage on the appliance

The electrical resistance of the thinner wire is more than the electrical resistance of the extension cord. According to Ohm’s law, for the same current in both the wire and cord, the value of voltage drop in the extension cord and in the thinner wire is to be reduced by this amount so that the value of current flow would be less that helps in the prevention the damage of the appliance.

The appliances might not work efficiently on the applied voltage.

Thus, the voltage to whatever appliance is used is reduced by this amount because the current flow through the appliance would be safer. The appliances might not work efficiently on the applied voltage.