An arc bridge has the following constants

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The dissipation factor can be calculated by using the following formula:

D = tanδ = (Zp - Zs) / Zs

where D is the dissipation factor, Zp is the impedance of the parallel arm, Zs is the impedance of the series arm, and tanδ is the tangent of the phase angle between the voltage and current.

Let's calculate the impedance of each arm:

Impedance of arm AB: Z_AB = R_AB - j/(ωC_AB) = 1kΩ - j318.3Ω

Impedance of arm AD: Z_AD = R_AD = 2kΩ

Impedance of arm BC: Z_BC = -j/(ωC_BC) = -j318.3Ω

Impedance of arm CD: Z_CD = R_CD + j/(ωC_x) = Rx + j(1/(ωC_x))

At 1 kHz, the angular frequency is ω = 2πf = 2π(1000) = 6283.2 rad/s.

Now, let's calculate the impedance of arm CD using the unknown capacitor and resistance: Z_CD = Rx + j(1/(6283.2*C_x))

When the bridge is balanced, the voltage between A and D is zero, which means that the currents flowing through arm AB and arm CD are equal. Therefore, the impedances of these two arms must be equal:

Z_AB = Z_CD

1kΩ - j318.3Ω = Rx + j(1/(6283.2*C_x))

Equating the real and imaginary parts separately, we get:

1kΩ = Rx

318.3Ω / (6283.2*C_x) = 1kΩ

C_x = 50.27 nF

Now, we can calculate the impedances of the parallel and series arms:

Zp = Z_AB || Z_BC = [(1kΩ)(-j318.3Ω)] / (1kΩ - j318.3Ω) = -j212.1Ω

Zs = Z_AD + Z_CD = 2kΩ + 1kΩ = 3kΩ

Finally, we can calculate the dissipation factor:

D = tanδ = (Zp - Zs) / Zs = (-j212.1Ω - 3kΩ) / 3kΩ = -0.071 - j0.071

The dissipation factor is approximately 0.1, which indicates a moderately low level of energy loss in the circuit.

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An arc bridge has the following constants

CBSE CLASS XII Related Questions

2.
This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?

4.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

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An arc bridge has the following constants

CBSE CLASS XII Related Questions

2.This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?

4.Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?

4.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.