In the circuit as following figure, the batteries have negligible internal resis

Question

In the circuit as following figure, the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 12 V. (a) Find the emf epsilon of the battery. (b) What will the ammeter read when the switch is closed? 3. A material of resistivity p is fanned into a solid, truncated cone of altitude L and radii a and b at either end. Calculate the resistance of the cone between the two flat end faces. 4. The region between two concentric conducting spheres with radii a and b is filled with a conducting material with resistivity p. When a potential difference is applied between the inner and outer surface, show the resistance is R = p/(4pi)*(1/a-1/b) 5. An uncharged capacitor and a resistor are connected in series to a battery as figure, where epsilon =5V, C = 2muF, and R=5times10^5ohm. The switch is thrown to position a. Find (a) the time constant of the circuit, (b) the maximum charge on the capacitor, (c) the maximum current in the circuit, and (d) the charge as functions of time and (e) current as function of time.

Explanation / Answer

a)

Current through 40 ohms or Ammeter reading is

IA=V/R =12/40 =0.3 A

Voltage across 10 ohms

V10 =IAR =0.3*10 =3 V

Voltage across 75 ohms is

V75 =V10+V40= 3+12 =15 Volt

From Voltage divider rule

V75=E*(75/75+30)

=>15 =E*(75/105)

E=21 Volts

b)

When switch is closed Ammeter reading is

I=20/40 =0.5 A

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