Electricity Worksheet8

An electric lamp, whose resistance is 20 Ω and a conductor of 4 Ω resistances are connected to a 6 V battery as shown in the figure. Calculate the total resistance of the circuit, the current through the circuit and the potential difference across the electric lamp and the conductor.

Three resistors of 2Ω, 3Ω and 4Ω are connected in (a) series (b) parallel. Find the equivalent resistance in each case.

Calculate the equivalent resistance between the points A and B in the circuits shown in figures.

A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 Ω, find the resistance of this combination.

A circuit consists of 1 Ω wire in series with a parallel arrangement of 6 Ω and 3 Ω wires. Calculate the total resistance of the circuit.

Calculate the effective resistances between the point A and B in the networks shown here.

Two resistors 4 Ω and 6 Ω are connected in parallel. The combination is connected across a 6 V battery of negligible resistance. Calculate (i) the current through the battery (ii) current through each resistor.

Carefully study the circuit diagram and calculate the value of x.

Three resistors of 6 Ω, 2 Ω and x are connected in series to a cell of emf 1.5V. the current registered is (1/6)A. calculate the value of x.

Calculate the equivalent resistances between the point A and B of the circuits shown here
Answer:

Since the lamp and conductor are in series,
total resistance in the circuit, R_{s} = 20 Ω + 4 Ω = 24 Ω
current through the circuit,
Potential difference across the lamp, V_{1} = IR_{1} = 0.25 A × 20 Ω = 5V
Potential difference across the conductor, V_{2} = IR_{2} = 0.25 A × 4Ω = 1V

(a) R_{s} = 2 Ω + 3 Ω + 4 Ω = 9 Ω
(b)

(a) Two resistances (6 Ω, 6 Ω) are in series and their resultant resistance = 6 Ω + 6 Ω = 12 Ω. Further, 12 Ω and 6 Ω are in parallel. The resultant resistance of 12 Ω and 6 Ω in parallel is .
(b) The resultant resistance of 6 Ω, 6 Ω which are in parallel in the first loop is and resultant resistance in the second loop is also 3 Ω. Since the two loops are in series, the total resistance between A and B is 3 Ω + 3 Ω = 6 Ω.

Resistance of a combination of 3 resistors (each of resistance 2 Ω) in series = 6 Ω
Resistance of 4 sets of such combinations in parallel .

Resistance of parallel combination of 6 Ω and 3Ω
Since 2 Ω and 1 Ω are in series, resultant resistance = 2 Ω + 1 Ω = 3 Ω.

1 Ω and 2 Ω are in series and their resultant resistance is . Now 3 Ω and 1.5 Ω are in parallel and their equivalent resistance
Resistance in the arm ACB, i.e., R_{1} = 5Ω; resistance in the arm ADB, i.e., R_{2} = 5Ω. Now, R_{1}, 1Ω and R_{2} are in parallel.
If R is their equivalent resistance, or

If R is the resultant resistance of 4 Ω and 6 Ω in parallel, then .
Current through the battery,
Since the resistors are in parallel, the potential difference
across each is the same, i.e., V.
Current through 4 Ω resistor
and current through 6 Ω resistance

The resistances 12 Ω, 6 Ω and 3 Ω are in parallel. If R_{1} is their resultant resistance, then
or
Since 2 Ω, R_{1} and x are in series,
Total resistance in the circuit, R = 2 + R_{1} + x = 2 + + x
From ohm’s law, or
0.4x = 4.514
x = 11.3 Ω

As
Or 8 + x = 9 or x = 1 Ω

(a) total resistance in the path ACB, i.e., R_{1} = 0.5 Ω + 2.5 Ω = 3 Ω
Total resistance in the path ADB, i.e.,
R_{2} = 8.5 Ω + 3.5 Ω = 12 Ω
Since R_{1} and R_{2} are in parallel, the effective resistance between the points A and B, i.e.,
(b) total resistance in the path ADCB, i.e., R_{1} = 8 + 9 + 7 = 24 Ω
Further, R_{1} (= 24 Ω) and 6 Ω are in parallel between the points A and B.
Thus, effective resistance between A and B, i.e.,