Question

Find the emf E1 in the circuit of the figure (Figure 1) .

Find the emf E2 in the circuit of the figure.

Find the potential difference of point b relative to point a

Answer 1

Concepts and reason

The concept required to solve this problem is Electromotive force, Ohm’s law, and Kirchhoff’s voltage law.

Initially, refer the circuit diagram given in the question. Later, find the values ${\varepsilon _1}$ and ${\varepsilon _2}$ of the circuit using the Kirchhoff’s voltage law. Finally, find the potential difference of the point b relative to the point a.

Fundamentals

The potential difference of point b relative to point a is as follows:

${V_{{\rm{ab}}}} = {V_{\rm{b}}} - {V_{\rm{a}}}$

Here, ${V_{\rm{b}}}$ is the potential of point b and ${V_{\rm{a}}}$ is the potential of point a.

According to the Kirchhoff’s voltage law, the sum of the voltages meeting at point a point is equal to zero.

The expression to write the voltage in the loop is equal to the the product of the current and the resistance of the resistor through which the current is flowing.

$V = IR$

Here, I is the current and R is the resistance.

(1)

The following figure shows the Kirchhoff’s voltage loop 1 and loop 2:

In the above figure, there loops are present. Loop 1 is the upper loop and loop 2 is the bigger loop.

According to the KVL law on the loop 1, the expression is as follows:

$20.{\rm{0 V}} - {\rm{1}}{\rm{.00 A}}\left( {1.00{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.00{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.00{\rm{ }}\Omega } \right) - {\varepsilon _1} - 6.00{\rm{ A}}\left( {1.00{\rm{ }}\Omega } \right) = 0$

Solve for ${\varepsilon _1}$ .

$\begin{array}{c}\\20{\rm{ V}} - {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.0{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) - {\varepsilon _1} - 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0\\\\20.{\rm{0 V}} - 1.00{\rm{ V}} + {\rm{4}}{\rm{.00 V}} + 1.{\rm{00 V}} - {\varepsilon _1} - 6.00{\rm{ V = 0}}\\\\{\varepsilon _1} = 18{\rm{ V}}\\\end{array}$

[(1)]

(1)

[(1)]

(2)

According to the KVL law on the loop, the expression is as follows:

$20{\rm{ V}} - {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) - 2.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) - 2.00{\rm{ A}}\left( {2.0{\rm{ }}\Omega } \right) - {\varepsilon _2} - 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0$

Solve for ${\varepsilon _2}$ .

$\begin{array}{c}\\20{\rm{ V}} - {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) - 2.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) - 2.00{\rm{ A}}\left( {2.0{\rm{ }}\Omega } \right) - {\varepsilon _2} - 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0\\\\20.{\rm{0 V}} - 1.00{\rm{ V}} - 2.{\rm{00 V}} - 4.{\rm{00 V}} - {\varepsilon _2} - 6.00{\rm{ V = 0}}\\\\{\varepsilon _2} = 7.0{\rm{ V}}\\\end{array}$

[(2)]

(2)

[(2)]

(3)

Apply the KVL law to the total loop and write the expression as follows:

${V_{\rm{b}}} + {\varepsilon _1} - 1.00{\rm{ A}}\left( {1.00{\rm{ }}\Omega } \right) - 4.00{\rm{ A}}\left( {1.00{\rm{ }}\Omega } \right) = {V_{\rm{a}}}$

Solve for ${V_{\rm{b}}} - {V_{\rm{a}}}$ .

${V_{\rm{b}}} - {V_{\rm{a}}} + {\varepsilon _1} - 1.00{\rm{ V}} - 4.00{\rm{ V}} = 0$

Substitute 18.0 V for ${\varepsilon _1}$ in the equation ${V_{\rm{b}}} - {V_{\rm{a}}} + {\varepsilon _1} - 1.00{\rm{ V}} - 4.00{\rm{ V}} = 0$ .

$\begin{array}{c}\\{V_{\rm{b}}} - {V_{\rm{a}}} + 18.0{\rm{ V}} - 1.00{\rm{ V}} - 4.00{\rm{ V}} = 0\\\\{V_{\rm{b}}} - {V_{\rm{a}}} + 13.0{\rm{ V = 0}}\\\\{V_{\rm{b}}} - {V_{\rm{a}}} = - 13.0{\rm{ V}}\\\end{array}$

[(3)]

(3)

[(3)]

Ans:

The magnitude of the ${\varepsilon _1}$ in the circuit of the figure is equal to 18.0 V.

The magnitude of the ${\varepsilon _2}$ in the circuit of the figure is equal to 7.00 V.

The magnitude of the potential difference of point b relative to point a is equal to -13.0 V.