Electricity Worksheet-15
(a) a 6 V battery in series with 1 Ω and 2 Ω resistors,
(b) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
Resistance of the resistor,
Current through the circuit,
In series, same current flows through all the resistors. Thus, current flowing through 12 Ω resistor
= 0.67 A
Resistance required in the circuit,
Resistance of each resistor r = 176 Ω
If n resistors, each of resistance r, are connected in parallel to get the required resistance R, then or or
To get a resistance of 9 Ω from three resistors, each of resistance 6 Ω, we connect two resistors in parallel and this parallel combination in series with the third resistor.
To get a resistance of 2 Ω from three resistors, each of resistance 6 Ω, we connect all the three resistors in parallel as equivalent resistance in parallel combination, i.e.,
R_{p} is given by
Resistance of each bulb,
Total resistance in the circuit,
Let n be the number of bulbs each of resistance r to be connected in parallel to get a resistance R.
110
resistance of each coil r = 24 Ω
(a) When each of the coils A or B is connected separately, current through each coil,
(b) When coils A and B are connected in series, equivalent resistance in the circuit,
Current through the series combination, i.e.,
(c) When the coils A and B are connected in parallel, equivalent resistance in the circuit,
Current through the parallel combination, i.e.,
18.3 A
Power used in 2 Ω resistor,
(ii) Since 4 V battery is in parallel with 12 Ω and 2 Ω resistors, potential difference across 2 Ω resistor, V = 4 V.
Power used in 2 Ω resistor,
Clearly,
resistance of the second lamp,
Now, the two lamps are connected in parallel, the equivalent resistance is given by
or
Current drawn from the line, i.e.,
Energy used by toaster in 10 min. (i.e., 1/6 h) = 1200 W × (1/6)h = 200 Wh
Hence a 250 W TV set uses more power in 1 h than a 1200 W toaster in 10 minutes.
Rate at which heat is developed, i.e., electric power,
P = I^{2} R = (15)^{2} × 8 = 1800 W = 1800 J/s