Question

three resistors are connected in parallel. the currents in the first three resistors 4mA,5mA,6mA respectively the voltage drop across tje fourth resistor is 200 volts the total power dissipated is 5watts. determine the value of resistance of the branches and the total resistance​

Answer 1

Answer:

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Explanation:

The circuit can be reduced, step by step, to a single equivalent resistance. The 8 ohm and the 8 ohm are connected in parallel , and so they can be replaced by an equivalent resistor Rp of 4 ohms, using the below equation.

Rp= (8+8)(8×8) =4ohms

This resistor is connected in series with the 4 ohm resistor R1. The total resistance Rt of the circuit is then

Rt =R1+Rp=4+4 = 8 ohms.

Since the 4 ohm and the 4 ohm are connected in series, they have the same current I1, which must be equal to the current of the battery. Using Ohms law we get,I= RR = 88 =1A.

Hence, the current flowing through the resistor R1 (4 ohms) is 1A.

Now the power dissipated in the resistor R1 (4 ohms) is given as follows.

P=I 2 R=1 2 ×4=1×4=4W.

Hence, the power dissipated in the resistor R1 (4 ohms) = 4W.

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Answer 2

Answer: Current = 25 mA and equivalent resistance is $\bold{2.99\times 10^{-5}}$ ohms.

Given:

• First three resistors 4 mA, 5 mA, 6 mA.
• Voltage drop across the fourth resistor is 200.
• Total power dissipated is 5 watts.

To Find: Value of resistance of the branches and the total resistance.

Explanation:

Step 1: When both of a resistor's terminals are linked to the respective terminals of another resistor or resistors, they are said to be connected in parallel.

But in a parallel resistive network, the voltage drop across each resistor is the same. Then, all parallel-connected elements, including resistors, have a common voltage applied across them.

Since there are four resistor and all are connected parallel to each other. Thus voltage across all the resistor are same i.e 200 V.

Power dissipated (P) = V.I

V = 200 V

P = 5 W

Thus, current = P / V = 5 / 200 = 0.025 A = 25 mA.

Resistance ($R_4$) = 200/0.025 = 8000 ohms

Resistance ($R_1$) = 200/0.004 = 50000 ohms

Resistance ($R_2$) = 200/0.005 = 40000 ohms

Resistance ($R_3$) = 200/0.006 = 33333.34 ohms

Hence value of resistances are 50000 ohms, 40000 ohms, 33333.34 ohms and 8000 ohms.

Step 2:

Total resistance is

$\frac{1}{R} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} +\frac{1}{R_4}\\\\\frac{1}{R} = \frac{1}{5000} +\frac{1}{40000} +\frac{1}{33333.34} +\frac{1}{8000}\\\\\frac{1}{R} = 33333.34\\\\R=2.99\times 10^{-5}$

Equivalent resistance is $\bold{2.99\times 10^{-5}}$ ohms.

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