1) 2) 3) 4) The circuit has been connected as shown in the figure for a \"long\"

Question

1)

2)

3)

4)

The circuit has been connected as shown in the figure for a "long"time. What is the magnitude of the electric potential epsilonC across the capacitor? If the battery is disconnected, how long does it take for the capacitor to discharge to epsilont/epsilon0 = 1/e of its initial voltage? In the figure below the battery has an emf of 18 V and an internal resistance of 1 ohamga. Assume there is a steady current flowing in the circuit. Find the charge on the 4 muF capacitor. Answer in units of muC Consider the resistor combination below. What is the resistance between a and b? Answer in units of Ohm A galvanometer deflects full scale for a current of 161 muA and has a resistance of 15 mu. A shunt resistor is added to this galvanometer to use it as a 30 A full-scale ammeter. What is the equivalent resistance of this galvanometer after it Is modified? Answer in units of ohm A 56 Ohm, 2.5 mA galvanometer is to be converted to a voltmeter that reads 78 V at full-scale deflection. What value of series resistance should be used with the galvanometer coil? Answer in units of Ohm

Explanation / Answer

capacitor will get fully charged and it will act as open circuit.

so current through 30 ohms and 34 ohms =32/(30+34)=0.5 A

so assuming voltage at -ve node of 32 volt source be 0 volts.

then voltage at connection of 30 ohms and 34 ohms=34*0.5=17 volts

current through 2 ohms and 62 ohms=32/(2+62)=0.5 A

voltage at connection of 2 ohms and 62 ohms=62*0.5=31 volts

so voltage across capacitor=31-17=14 volts

option 1 is correct.

Q2.time constant is R*C.

where R=net resistance=(30 ohms and 2 ohms in series) in parallel with (34 ohms and 62 ohms in series)

=32 ohms in parallel with 96 ohms

=32*96/(32+96)

=24 ohms

so time constant=24*C=0.36 ms

so option 6 is correct.

Q3.

after fully charged,

capacitor will act like open circuit.

so current in circuit=18/(1+8+8)=1.0588 A

voltage across capacitor=voltage across 8 ohms=8*1.0588=8.47 volts

charge=C*V=33.88 uC

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