Difference Between EMF and Voltage: Definitions & Examples

Ans: V = 3.2V, Ir = 0.5 ohms, I = 0.6 A

Applying the EMF Formula:

∈ = V+Ir

=3.2 V+(0.6 A)(0.5 Ω)

=3.5 V

So the emf of the circuit is 3.5 V

Ans: The force that causes electrons to flow through a current in an electrical circuit is represented by the voltage of a battery. It measures potential energy, which is the amount of energy that may be used to transfer electrons from one place in the circuit to another. An opposing force called resistance might obstruct the actual flow of electrons through the circuit. The current multiplied by the resistance equals the voltage present. Voltage (E) = Current (I) x Resistance (R), or E = IR, is the formula.

Ans: Given,

Potential difference = V = 4V

Current in the circuit = I =0.7A

The Internal resistance of the battery is r = 0.8

Now, Emf of the circuit is:

E = I(R+r)

Where,

R- External resistance of the electrical circuit.

r- Internal resistance of the given circuit

I- Current flowing through the circuit

On rearranging the above expression,

E = IR+Ir

We know that the potential difference across the resistance is equal to the product of the current in the circuit and the external resistance. Thus,

E = V+Ir

Substituting given values in the equation,

E = 4 + (0.7 x 0.8) = 4.56 volts

Therefore, the EMF of the battery is given by 4.56 V.

Ans: When a circuit is turned on, the potential difference across the terminals of a load is called terminal voltage. The largest potential difference that a cell or generator can deliver when no current flows through it is called E.m.f. The terminal voltage is always less than the cell's emf.

Ans: Because EMF is not a force, but rather a "potential" to provide energy, the term "force" is a little misleading. Because of its historical significance, the term EMF is still used to distinguish between voltages generated and energy lost through resistors.

Ans: The electromotive force (EMF) is the potential difference between the terminals of a cell when no current is extracted from it (or e.m.f.)

The voltage at the terminal:- The terminal voltage of a cell is the potential difference between its electrodes when current is extracted from it.

Voltage drop: Current passes through the circuit while the electric cell is in a closed circuit, resulting in a voltage drop. There is a potential drop across the cell's internal resistance. As a result, the potential drop across the terminals in a closed circuit is lower by an amount equal to the potential drop across the internal resistance of the cell.

Ans: When a circuit is turned on, the potential difference across the terminals of a load is called terminal voltage.

The largest potential difference that a cell or generator can deliver when no current flows through it is called E.m.f. A voltmeter is used to measure the terminal voltage.

A potentiometer is used to measure the e.m.f.

The terminal voltage is always less than the cell's emf. It's due to the current going through the cell's or generator's internal resistance, which causes a drop in potential. The following equation connects the emf (E) with the terminal (V).

E = V + Ir, where I is the current and r is the cell's internal resistance.

Ans: When the cell is being charged.

The formula will be: Terminal voltage= E+ir, where E+ir>E

Ans: For parallel connection of n identical cells, the net emf is equal to that of one cell and the net internal resistance is the parallel combination of resistances of all the cells \(r_{eq} = \frac{r}{n}\)

Four cells are connected in parallel to the parallel combination of two 15 Ω resistors as shown below.

Let r be the internal resistance of each cell and I will be the current I in the circuit. Since the cells are connected in parallel. Total e.m.f in the circuit – e.m.f. one cell -2V Further, the total internal resistance of the cells is given by:

Let R be the resistance of the parallel combination of two 15 Ω resistors. Then, total external resistance,

\(R= \frac{15 \times 15}{15 + 15}.75 \Omega\)

Now, the internal resistance of the parallel combination of cells is given by:

Ans: V_R = V_m sin wt; V_Y = V_m sin (wt-120⁰); V_B = V_m sin (wt-240⁰)

Ans: E1 = 4.44*N1* f* Bm*A and E2 = 4.44*N2*f*Bm*A

Ans: Significance of back EMF is as follows:

• When the motor is running on no load, small torque is required to overcome the friction and windage losses. Therefore, the armature current Ia is small and the back emf is nearly equal to the applied voltage.
• If the motor is suddenly loaded, the first effect is to cause the armature to slow down. Therefore, the speed at which the armature conductors move through the field is reduced and hence the back emfEb falls. The decreased back emf allows a larger current to flow through the armature and larger current means increased driving torque. Thus, the driving torque increases as the motor slows down. The motor will stop slowing down when the armature current is just sufficient to produce the increased torque required by the load.
• If the load on the motor is decreased, the driving torque is momentarily in excess of the requirement so that armature is accelerated. As the armature speed increases, the back emf Eb also increases and causes the armature current Ia to decrease. The motor will stop accelerating when the armature current is just sufficient to produce the reduced torque required by the load

Ans: Use formulae for series and parallel combinations of resistances. P.D. across each branch is given by Ohm’s law.

While the slide is in the middle of the potentiometer only half of its resistance (R₀/2) will be between points A and B. Hence, The total resistance between A and B, say R1 will be given by the following expression:

Thus, the main current from the battery is:

\(i = \frac{100V}{400 \Omega}= 0.25A\)

The potential drop across the 200 Ω resistance is, therefore, 200Ω × 0.25 A = 50 V and that across 300 is also 50 V. This is also the potential drop across the voltmeter, and hence the reading of the voltmeter is 50 V.

Ans: Resonance is defined as a phenomenon in which applied voltage and resulting current are in phase. In other words, an AC circuit is said to be in resonance if it exhibits a unity power factor condition, which means applied voltage and resulting current are in phase.

Ans: The Apparent power (in VA) is the product of the rms values of voltage and current. S = V_rms I_rms

The Power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the load impedance. And Power factor = cos φ

The pf is lagging if the current lags voltage (inductive load) and is leading when the current leads voltage(capacitive load).