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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 “Find the emf e1 in the circuit of the figure (figure 1) …”

Concepts

This problem can be solved using the concepts of Ohm’s law and Kirchhoff’s voltage laws, Electromotive force, Ohms law, or Kirchhoff’s power law.

Initially, refer to the circuit diagram in the question. Later, find the values ε1​ and ε2​ of the circuit using the Kirchhoff’s voltage law. Finally, find the difference between the point b and the point a.

Fundamentals

The potential difference of point b relative to point a is as follows: Vab​=Vb​−Va​

Here Vb refers to the potential for point b, and Va refers to the potential for point a.

The Kirchhoff’s voltage law states that the sum of all voltages at point a is zero.

To write the voltage inside the loop, you need to multiply the current and resistor by the sum of their resistance.

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:

20.0V−1.00A(1.00Ω)+1.00A(4.00Ω)+1.00A(4.00Ω)−ε1​−6.00A(1.00Ω)=0

Solve for ε1​ 

20V−1.00A(1.0Ω)+1.00A(4.0Ω)+1.00A(1.0Ω)−ε1​−6.00A(1.0Ω)=0

20.0V−1.00V+4.00V+1.00V−ε1​−6.00V=0

ε1​=18V​

[(1)]

(1)

[(1)]

(2)

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

20V−1.00A(1.0Ω)−2.00A(1.0Ω)−2.00A(2.0Ω)−ε2​−6.00A(1.0Ω)=0

Solve for ε2​ .

20V−1.00A(1.0Ω)−2.00A(1.0Ω)−2.00A(2.0Ω)−ε2​−6.00A(1.0Ω)=0

20.0V−1.00V−2.00V−4.00V−ε2​−6.00V=0

ε2​=7.0V​

[(2)]

(2)

[(2)]

(3)

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

Vb​+ε1​−1.00A(1.00Ω)−4.00A(1.00Ω)=Va​

Solve for Vb​−Va​ .

Vb​−Va​+ε1​−1.00V−4.00V=0

Substitute 18.0 V for ε1​ in the equation Vb​−Va​+ε1​−1.00V−4.00V=0 .

Vb​−Va​+18.0V−1.00V−4.00V=0

Vb​−Va​+13.0V=0

Vb​−Va​=−13.0V​

[(3)]

(3)

[(3)]

Ans:

• The magnitude of the ε1​ in the circuit of the figure is equal to 18.0 V.
• The magnitude of the ε2​ 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.

Last words

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