Answer to Question #92721 in Electric Circuits for Efache

Question

A battery of four cells each of e.m.f. 1.5V and internal resistance 1.0 ohm are connected to a 1 ohm resistor in series with a parallel combination of two 2 ohms resistors. Draw the circuit diagram and calculate:
(a) the combined external resistance.
(b) the current in the circuit.
(c) the lost volts in the battery.
(d) the current in one of the two resistors

Answer 1

Part a

The parallel resistance is calculated.

$R_{parallel}=\frac{2.00\Omega *2.00\Omega}{2.00\Omega+2.00\Omega}=\frac{4.00\Omega^{2}}{4.00\Omega}=1.00\Omega$

Now the equivalent resistance is added in series resistance.

Assessing numerically.

$R_{external}=R_{Parallel}+1.00\Omega$

$R_{external}=1.00\Omega+1.00\Omega=2.00\Omega$

Finally:

$R_{external}=2.00\Omega$

Part b

Considering the internal resistance, the equivalent resistance is:

$R_{eq}=R_{external}+R_{Sourse}=2.00\Omega+1.00\Omega=3.00\Omega$

Applying Ohm's Law.

$I_{T}=\frac{V}{R_{eq}}$

Where:

• Source Potential Difference

$V=1.5V$

• Total resistance of the circuit.

$R_{eq}=3.00\Omega$

Numerically evaluating

$I_{T}=\frac{1.5V}{3.00\Omega}=0.5A$

The total current in the circuit is: $I_{T}=0.5A$

Part c

The voltage drop in the resistor due to internal resistance is calculated by Ohm's Law.

$V=I_{T}*R_{sourse}$

Where:

• Resistance value of the source.

$R_{sourse}=1.00\Omega$

• Electric current

$I_{T}=0.5A$

Numerically evaluating.

$V=0.5A*1.00\Omega=0.5V$

The voltage drop at the source is:

$V=0.5V$

Part d

First, the tension in the branch is calculated in parallel using Ohm's Law.

$V=I_{T}*R_{parallel}$

Where:

• Total current

$R_{Parallel}=1.00\Omega$

• parallel resistance.

$I_{T}=0.5A$

Evaluating numerically

$V=0.5A*1.00\Omega=0.5V$

Now using Ohm's Law, the current in one of the resistors is calculated.

$I=\frac{V_{parallel}}{R}$

Where:

• Potential difference

$V_{parallel}=0.5V$

• resistance

$R=2.00\Omega$

Evaluating numerically.

$I=\frac{0.5V}{2.00\Omega}=0.25A$

The current through one of the two resistors is:

$I=0.25A$