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 votage 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 0.55 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 ofthe battery 8. Two unknown resistors connected in series with a battery, dissipate 625 w when drawing a total current of 5.0 A. The dissipate only 100 W. (a) First resistor is (b) Second resistor is .. capac 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 0.72 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) current is reduced to 1/3, it means resistance becomes thrice.

(R+400)/R = 3

R + 400 = 3R, or R = 200 ohm answer

b) final voltage drop across R, V = 12*200/(200+400) = 4 V answer

7] a) Let emf be E and internal resistance be r.

we know that terminal voltage V = E - i*r

5 = E - 0.55r

and 5.7 = E - 0.18r

subtracting first equation from second, 0.7 = 0.37r

r = 0.7/0.37 = 1.892 ohm

E = 5.7 +0.18*1.892 = 6.04 V answer

b] internal resistance = 1.892 ohm, found above

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