Calculate the power dissipated in the 2 Ohm resistor, if the voltage of the batt

Question

Calculate the power dissipated in the 2 Ohm resistor, if the voltage of the battery is epsilon 21 V and the resistance of the resistor is R = 4.5 Ohms? Express the answer to two decimal places. Find the resistance of a 23 meter length of metal wire of 0.4 mm diameter. The resistivity of metal is 1.7 times 10^-6 Ohm cm at a temperature of 20 degree C. Express the answer with two decimal places. In the circuit shown in Figure, an ideal ohmmeter is connected across ab with the switch S open. All the connecting leads have negligible resistance. What will be the ohmmeter reading in Ohms (with two decimal places) if R_1 = 7.4 Ohm; R_2 = 8.30 and R_3 6-8 Ohm?

Explanation / Answer

here,

1)

E = 21 V

R = 4.5 ohm

(1 + 2) ohm and (5 + 1) ohm are in parallel, Req = ( 5 + 1) * ( 1 + 2) /( 5 + 1 + 1 + 2)

R' = 2.33 ohm

the current through R ohm , I = E /( R + R')

I = 21 /( R' + R) = 21 /( 2.33 + 4.5)

I = 3.07 A

the current through 2 ohm , I' = I * ( 5 + 1) /( 2 + 1+ 1 + 5) A

I' = 3.07 * ( 5 + 1) /( 2 + 1+ 1 + 5) A

I' = 2.05 A

the power dissapated through 2 ohm , P = I'^2 * 2 = 2.05^2 * 2 = 8.4 W

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