6. The current through a resistor R is reduced to one-third of its initial value

Question

6. The current through a resistor R is reduced to one-third of its initial value when a 400 resistor is added in series with it. (a) Calculate the value of R voltage drop across R if the system is connected to a 12 V battery (b) Calculate the final 7. The terminal voltage of a battery is 5.0 V when 0.5S A current is drawn from it, whereas the terminal voltage is 5.7 V when a current of 180 mA is drawn. (a) Calculate the emf of the battery (b) What is the internal resistance of the battery 8. Two unknown resistors connected in series with a battery, dissipate 625 W when drawing a total current of 5.0 A. The same resistors connected in parallel and with same total current (50A) (b) Second resistor is 9, A 6.0 V battery is being used to charge a 5.0 F capacitor through a resistor R. The capacitor attains a potential difference of 3.6 V in 1.0 s after the charging begins. (a) Calculate the value of (b) Calculate the current through the resistor at-1.0s 10. A capacitor connected in series with a resistor R is being charged by a power source. The capacitor charges to 55% of its maximum value in 072 ms.(a) Calculate the time constant of the circuit (b) If the capacitor has a value 7.0 F, what is the value of the resistor R

Explanation / Answer

6] a]

Let i be the initial current flowing through resistor R.

New resistance = R + 400

and new current I = i/3

voltage across the circuit will be the same

so, iR = (i/3)[R + 400]

=> 3R = R + 400

=> R = 200 ohms

b] Total resistance = R' = 200 + 400 = 600 ohms

so, for a 12 volts battery, the current in the circuit will be:

I = V/600 = 12/600 = 0.02 A

therefore, the voltage drop across R will be:

v = IR = 0.02(200) = 4 volts.

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