19. In an AC circuit with all components connected in
series, the emf and the current are given by
TC
V=200 sin 314t +
volt and I=5 sin 314t ampere.
6
중
27.
Obtain (1) the peak values of emf and current, (i) the
frequency of the AC source, (iii) the phase difference
between V and I (iv) impedance of the circuit
Answers
Answer:
19. In an AC circuit with all components connected in
series, the emf and the current are given by
TC
V=200 sin 314t +
volt and I=5 sin 314t ampere.
6
중
27.
Obtain (1) the peak values of emf and current, (i) the
frequency of the AC source, (iii) the phase difference
between V and I (iv) impedance of the circuit
Explanation:
don't no
Given:
To find:
- Peak value of EMF and current.
- Frequency of the AC source.
- Phase difference between V and I.
- Impedance of the circuit.
Solution:
Step 1
Standard form of equation is ω ; where is the peak value.
We have been given the equations for a variable EMF and current with respect to time t.
Hence,
Peak value of EMF =
Peak value of Current =
Step 2
If we consider the standard form of an equation for current,
(ω)
Comparing it with the given equation,
We get ω
We know, ω = 2πf ; where f is the frequency
Hence,
πf
Hence, frequency is
Step 3
We know, if in a circuit, only resistor is present, then the current and voltage in the circuit work in the same phase.
Hence, Phase difference between V and I is 0.
Step 4
We know,
Impedance in a circuit is given by , here, in the circuit, Inductor and capacitor is not present hence, the impedance in the circuit will be solely the resistance in the circuit.
Using Ohm's law,
Hence, or
Substituting the given values, we get
Ω
Hence, the impedance of the circuit is 40Ω.
Final answer:
Hence,
- Peak value of current is 5A and peak value of EMF is 200V.
- Frequency of the AC source is 50 Hz.
- Phase difference between V and I is .
- Impedance of the circuit is 40Ω.