Series circuit with meters

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The idea

Two resistors hang one behind the other on a battery. You measure the current at two places and the voltage across each resistor. What do you notice? The current is the same everywhere, the voltage is shared among the resistors. You already know how a series circuit is built; now the quantities come in.

What to remember

In a series circuit: - The current is the same everywhere: $I = I_{1} = I_{2}$ . - The voltages add up: $U = U_{1} + U_{2}$ . - The resistances add up: $R = R_{1} + R_{2}$ .

Total resistance: The total resistance is the resistance the whole circuit presents to the source. In a series circuit it is the sum of the single resistances.

I = \frac{U}{R_{1} + R_{2}}

A rectangle of wires with a battery on the left, two rectangles for resistors at the top and a circle with A on the right. A circle with V is connected across the first resistor. — Two resistors in series. The ammeter sits in the circuit, the voltmeter in parallel with the first resistor. Labels: 1 battery, 2 first resistor, 3 second resistor, 4 ammeter, 5 voltmeter.
Source: https://github.com/lernapps/elektrizitaet. Licence: CC0-1.0.

A larger resistor takes more voltage

The larger resistor also gets the larger share of the voltage. The current through both is the same.

Example

Two resistors on 12 V

A resistor of 10 Ω and one of 30 Ω are connected in series on 12 V. The total resistance is $10 + 30 = 40$ Ω. The current is $I = \frac{12 V}{40 Ω} = 0.3$ A, everywhere in the circuit. Across the 10 Ω resistor there are $0.3 \cdot 10 = 3$ V, across the 30 Ω resistor $0.3 \cdot 30 = 9$ V. Together that is 12 V.

Practice

Task no. 1

Two resistors of 10 Ω and 30 Ω are connected in series on 9 V. What does the voltmeter show across the second resistor (30 Ω)?

The circuit is switched off. Predict the number, then switch on: the voltmeter is across the second resistor.

Solution

The current is 9 V : (10 + 30) Ω; across the second resistor there are 30 Ω · current = 6.75 V.

Exercise

Three worked tasks, each with its solution:

Task no. 1
Two resistors of 10 Ω and 10 Ω are connected in series on 9 V. What is the current in amperes?
Hint
First calculate the total resistance $R_{1} + R_{2}$ .
Solution
1. $R = R_{1} + R_{2} = 10 Ω + 10 Ω$
2. $I = \frac{U}{R} = \frac{9 V}{10 + 10 Ω} = 0.45 A$
Right answer: 0.45 A
Task no. 2
Two resistors of 50 Ω and 50 Ω are connected in series on 6 V. What is the current in amperes?
Hint
First calculate the total resistance $R_{1} + R_{2}$ .
Solution
1. $R = R_{1} + R_{2} = 50 Ω + 50 Ω$
2. $I = \frac{U}{R} = \frac{6 V}{50 + 50 Ω} = 0.06 A$
Right answer: 0.06 A
Task no. 3
Two resistors of 20 Ω and 10 Ω are connected in series on 9 V. What is the current in amperes?
Hint
First calculate the total resistance $R_{1} + R_{2}$ .
Solution
1. $R = R_{1} + R_{2} = 20 Ω + 10 Ω$
2. $I = \frac{U}{R} = \frac{9 V}{20 + 10 Ω} = 0.3 A$
Right answer: 0.3 A

Exercise

Three worked tasks, each with its solution:

Task no. 1
Two resistors are connected in series on 6 V. The voltmeter at the first resistor shows 2 V. What is the voltage across the second resistor in volts?
Hint
$U = U_{1} + U_{2}$ , so $U_{2} = U - U_{1}$ .
Solution
1. $U_{2} = U - U_{1} = 6 V - 2 V = 4 V$
Right answer: 4 V
Task no. 2
Two resistors are connected in series on 9 V. The voltmeter at the first resistor shows 4 V. What is the voltage across the second resistor in volts?
Hint
$U = U_{1} + U_{2}$ , so $U_{2} = U - U_{1}$ .
Solution
1. $U_{2} = U - U_{1} = 9 V - 4 V = 5 V$
Right answer: 5 V
Task no. 3
Two resistors are connected in series on 9 V. The voltmeter at the first resistor shows 1 V. What is the voltage across the second resistor in volts?
Hint
$U = U_{1} + U_{2}$ , so $U_{2} = U - U_{1}$ .
Solution
1. $U_{2} = U - U_{1} = 9 V - 1 V = 8 V$
Right answer: 8 V

Exercise

Solution

0.2 A right answer
In a series circuit there is only one current, and it is the same everywhere.
less than 0.2 A
The current is not used up in the resistors. As much comes out as flows in.
0.4 A
The current does not double. Behind the second resistor no more flows than before.

In a series circuit the current is the same at every place.