# Series Circuits, Resistors in series & equivalent resistance – numerical

What is a series circuit? A series circuit consists of several resistors connected in sequence, one after the other. This increases the length of the conductor. Therefore total resistance increases as the number of resistors in series are increased in the circuit.

## Characteristics of Series Circuit as found from experiments

Experiments show that the current in the series circuit made up one loop is the same everywhere.

As there are potential drops across each resistor, the total voltage across the battery is equal to the sum of the potential drops across each resistor.

If one resistor goes out, then the whole circuit goes off. (like when a light bulb in series circuit goes out, the entire circuit becomes inoperative)

## Equivalent Resistance of a series circuit

Using Ohm’s Law and our experimental results, we find that: (see the circuit diagram above)

The current flowing through the circuit = I = current through each of the resistors in series
I = I1 = I2 = I3
VT = Total voltage or Potential difference
Req = VT/I = equivalent resistance

VT = V1 + V2 + V3 = sum of potential drops across each resistor

Req =  VT/I = (V1 + V2 + V3) /I = V1/I + V2/I + V3/I = V1/I1 + V2/I2 + V3/I3 = R1 + R2 + R3

Req = R1 + R2 + R3

## Numerical Problems (based on series circuit and its equivalent resistance)

1] Resistors of 10 Ω, 15 Ω, and 5 Ω are all connected in series to a 90 V source of potential difference. Calculate the total equivalent resistance of the circuit, the total current in the circuit, and the potential drop across each resistor.

Solution:
Equivalent Resistance = Req = (10 + 15 + 5) Ω = 30 Ω
Total current in the circuit I = V/Req = 90 V/ 30 Ω = 3 Ampere

Potential drop across each resistor:
Potential drop across 10 Ω resistor = I x resistance = 3 x 10 V = 30 V

Potential drop across 15 Ω resistor = I x resistance = 3 x 15 V = 45 V

Potential drop across 5 Ω resistor = I x resistance = 3 x 5 V = 15 V

2] Three resistors of 50 Ω, 30 Ω, and 40 Ω are connected in series to a 60 V battery. Calculate the equivalent resistance of the circuit, the total current in the circuit, the total power generated by the circuit, and the voltage drop across the resistors.

Solution:
Equivalent Resistance = (50 + 30 + 40) Ω = 120 Ω
Total current = I = V/R = 60/120 Ampere = 0.5 A [ we used Ohm’s law here]
Total power = P = VI = 60 x 0.5 Watt = 30 W
Voltage drop across 50 Ω = IR =0.5 x 50 V= 25 V
Voltage drop across 30 Ω = IR = 0.5 x 30 V = 15 V
Voltage drop across 40 Ω = IR = 0.5 x 40 V = 20 V

Formula Used

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Ohm’s Law Numericals

Series Circuits, Resistors in series & equivalent resistance – numerical
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