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# Internal Resistance

## The voltage provided by a battery is less than the label says due to the parts of the battery charge must flow through.

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Internal Resistance

A battery is a voltage source. A battery can be thought of as a perfect voltage source with a small resistor (called internal resistance) in series. The electric energy density produced by the chemistry of the battery is called emf, but the amount of voltage available from the battery is called terminal voltage. The terminal voltage equals the emf minus the voltage drop across the internal resistance (current of the external circuit times the internal resistance). In practice, if you short circuit a battery and measure its voltage you will see the voltage is less than what is marked on it and what it can produce when outputting smaller currents. The short circuit of the battery, makes it pump out a lot of current and then the voltage drop over the internal resistance gets large ( V = Ir ) which in turn reduces the terminal voltage.

Key Equations

Vterminal=emfIr\begin{align*} V_{terminal} = emf - Ir \end{align*}

The terminal voltage (or 'output voltage') is equal to the emf (it's 'ideal voltage') minus the voltage drop across the internal resistance.

#### Example

You have a battery with an emf of 5 V and an unknown internal resistance. You hook the battery up to a circuit with one 3Ω\begin{align*}3\;\Omega\end{align*} resistor and measure the current through the resistor to be 1.5 A. What is the internal resistance of the batter and how much power is the battery's resistance dissipating.

To start this problem we'll first find the terminal voltage of the battery using the information we know about the resistor. We know the voltage drop across the resistor must be equal to the terminal voltage because there the total change in voltage must be over the whole circuit.

VVV=IR=1.5A3Ω=4.5V\begin{align*} V&=IR\\ V&=1.5\:\text{A}*3\:\Omega\\ V&=4.5\:\text{V}\\ \end{align*}

We can plug this value into the equation given above to find the internal resistance.

V=emfIrr=emfVIr=5V4.5V1.5Ar=.33Ω\begin{align*} V=emf-Ir\\ r=\frac{emf-V}{I}\\ r=\frac{5\:\text{V}-4.5\:\text{V}}{1.5\:\text{A}}\\ r=.33\:\Omega\\ \end{align*}

Now we can find find the power dissipated by the resistor

PPPPP=IV=I(Ir)=I2r=(1.5A)2.33Ω=.75W\begin{align*} P&=IV\\ P&=I(Ir)\\ P&=I^2r\\ P&=(1.5\:\text{A})^2*.33\:\Omega\\ P&=.75\:\text{W}\\ \end{align*}

### Review

1. In the circuit shown here, the battery produces an emf of 1.5V\begin{align*}1.5\;\mathrm{V}\end{align*} and has an internal resistance of 0.5 Ω\begin{align*}0.5\ \Omega\end{align*}.
1. Find the total resistance of the external circuit.
2. Find the current drawn from the battery.
3. Determine the terminal voltage of the battery
4. Show the proper connection of an ammeter and a voltmeter that could measure voltage across and current through the 2 Ω\begin{align*}2\ \Omega\end{align*} resistor. What measurements would these instruments read?
2. Students are now measuring the terminal voltage of a battery hooked up to an external circuit. They change the external circuit four times and develop the Table (below); using this data, complete the following:
1. Graph this data, with the voltage on the vertical axis.
2. Use the graph to determine the emf of the battery.
3. Use the graph to determine the internal resistance of the battery.
4. What voltage would the battery read if it were not hooked up to an external circuit?
Terminal Voltage (v)\begin{align*}(v)\end{align*} Current (a)\begin{align*}(a)\end{align*}
14.63\begin{align*}14.63\end{align*} .15\begin{align*}.15\end{align*}
14.13\begin{align*}14.13\end{align*} .35\begin{align*}.35\end{align*}
13.62\begin{align*}13.62\end{align*} .55\begin{align*}.55\end{align*}
12.88\begin{align*}12.88\end{align*} .85\begin{align*}.85\end{align*}
3. You have a battery with an emf of 12V\begin{align*}12\;\mathrm{V}\end{align*} and an internal resistance of 1.00 Ω\begin{align*}1.00\ \Omega\end{align*}. Some 2.00A\begin{align*}2.00\;\mathrm{A}\end{align*}are drawn from the external circuit.
1. What is the terminal voltage
2. The external circuit consists of device X\begin{align*}X\end{align*}, 0.5A\begin{align*}0.5\;\mathrm{A}\end{align*} and 6V\begin{align*}6\;\mathrm{V}\end{align*}; device Y\begin{align*}Y\end{align*}, 0.5A\begin{align*}0.5\;\mathrm{A}\end{align*} and 10V\begin{align*}10\;\mathrm{V}\end{align*}, and two different resistors. Show how this circuit is connected.
3. Determine the values of the two resistors.

1. a. 3.66 Ω\begin{align*}3.66 \ \Omega\end{align*} b. 0.36A\begin{align*}0.36\mathrm{A}\end{align*} c. 1.32V\begin{align*}1.32 \;\mathrm{V}\end{align*} d. voltmeter: 0.252 V; ammeter: 0.126 A
2. b. 15V\begin{align*}15\text{V}\end{align*} c. 2.5Ω\begin{align*}2.5\Omega\end{align*} d. 15V\begin{align*}15\text{V}\end{align*}
3. a.10V\begin{align*}10\text{V}\end{align*} b. c. R1=2.6Ω\begin{align*}R_1=2.6\Omega\end{align*} and R2=6Ω\begin{align*}R_2=6\Omega\end{align*}

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