Advertisements
Advertisements
प्रश्न
A battery of emf 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series.
Find:
1) Current in the circuit
2) The terminal voltage of the cell
3) The potential difference across 6Ω Resistor
4) Electrical energy spent per minute in the 4Ω resistor.
Advertisements
उत्तर
E = 12 V; ri = 2 Ω; RA = 4 Ω; RB = 6 Ω
1) The current in the circuit is
`I = E/R_"total" = E/(r_i + r_A + r_B)`
`:. I = 12/(2+4+6) = 1 A`
2) The terminal voltage of the cell is
V = E - Iri
∴ V = 12 - (1 x 2) = 12 - 2 = 10 V
3) The potential difference across the 6 Ω is
VB = IRB
∴ VB = 1 x 6 = 6V
4) The electrical energy spent per minute (=60 s) is
E = I2Rt
E = 12 X 4 X 60 = 240J
APPEARS IN
संबंधित प्रश्न
By what other name is the unit joule/coulomb called?
The p.d. across a lamp is 12 V. How many joules of electrical energy are changed into heat and light when:
a charge of 5 C passes through it?
A current of 5 amperes flows through a wire whose ends are at a potential difference of 3 volts. Calculate the resistance of the wire.
Explain the analogy between the flow of charge (or current) in a conductor under a potential difference with the free fall of a body under gravity.
The following table shows current in Amperes and potential difference in Volts.
What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
Calculate the current flowing through each of the resistors A and B in the circuit shown in the following figure.
State Ohm’s law.
If P and V are the power and potential of device, the power consumed with a supply potential V1 is:
Two charged spherical conductors of radius R1 and R2 are connected by a wire. Then the ratio of surface charge densities of the spheres `(σ_1//σ_2 )` is ______.
Twenty-seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
