A step-up transformer for long-range transmission of electric power is used to c

Question

A step-up transformer for long-range transmission of electric power is used to create a potential difference of 1.2768 times 10^5 V across the secondary. The potential difference across the primary is 199 V and the secondary has 29300 turns. How many turns are in the primary? Answer in units of turns. A television picture tube requires a high potential difference, which in older models is provided by a step-up transformer. The transformer has 19 turns in its primary and 2921 turns in its secondary. A potential difference of 163 V is placed across the primary. What is the output potential difference? Answer in units of V. The alternating potential difference of a generator is represented by the equation epsilon = (308 V)sin(431 rad/s) t, where epsilon is in volts and l is in seconds. Kind the frequency of the potential difference of the source. Answer in units of Hz. Find maximum potential difference output of the source. Answer in units of V. An RLC circuit consists of a 396 Ohm resistor, a 77 mu F capacitor, and a 326 mH inductor, connected in series with a 69.9 V, 99.9 Hz power supply. What is the phase angle between the current and the applied voltage? Answer between - 180 degree and + 180 degree. Answer in units of^compositefunction. A transformer operating from 74 V (rms supplies a 8.3 V lighting system for a garden. Eight lights, each rated 53 W, are installed parallel. Find the equivalent resistance of the system. Answer in units of Ohm. What is the current in the secondary circuit? Answer in units of A. What single resistance, connected across the 74 V supply, would consume the same power as when the transformer is used? Answer in units of Ohm. A series RLC circuit with L = 7.5 mH, C = 5 mu F, and R = 2.5 Ohm is driven by a generator with a maximum emf of 120 V and a variable angular frequency omega. Find the resonant frequency omega_0. Answer in units of rad/s. Find the at resonance. Answer in units of A. Find X_C when omega = 10000 rad/s. Answer in units of Ohm. Find X_L when omega = 10000 rad/s. Answer in units of Ohm. Find Z when omega = 10000 rad/s. Answer in units of Ohm. Find I_rms when omega = 10000 rad/s. Answer in units of A. Find the phase angle when omega = 10000 rad/s. Answer in units of^compositefunction.

Explanation / Answer

Given : V = 74 v ; V2= 8.3 V ; P = 53 W

Part :1 Solution -

Since lights are installed in parallel, the voltage across each is same.

i.e. For each light , P = v22/R

R = V22/P = (8.3)2/53 = 1.3 ohm

For the 8 lights

1/Req= 8/R

Req= 1.3/8 = 0.162 ohm

Part:2 solution -

current in the secondary circuit can be given as:

I2= v2/Req= 8.3/0.162 = 51.08 A

Part:3 solution -

Total power consumed in secondary circuit is :

P2= 8×74v = 592 W

We need to find R1 sucsuch that

P2= I12R1 [ I1 is the current in primary]

I1= I2V2/V1

= (51.08)(8.3)/74 = 5.81 A

Hence required resistance is

R1= P2/I12= 592/(5.81)2= 17.54 ohm

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