Explain Ohm's law and calculate the EMF and internal resistance of a battery from a V-I graph.

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8.1.1

Ohm's law states that the current flowing through a conductor between two points is directly proportional to the potential difference across the two points, and inversely proportional to the resistance between them, provided the temperature and other physical conditions remain constant.

8.1.2

You need to draw a straight line of best fit through the plotted points on the graph. Ensure this line extends to cut both the potential difference (y-axis) and current (x-axis). For the calculations below, we will assume such a line has been drawn.

8.1.3

The EMF ( $\mathcal{E}$ ) of the battery is the terminal potential difference when the current ( $I$ ) is zero. This corresponds to the y-intercept of the graph.
From the graph, the y-intercept is $5.5 \, \text{V}$ .
Therefore, the EMF of the battery is:

\mathcal{E} = \boxed{\text{5.5 V}}

8.1.4

The internal resistance ( $r$ ) of the battery is the negative of the gradient of the potential difference versus current graph ( $V = \mathcal{E} - Ir$ ).

Identify two points from the line of best fit.
We'll use the intercepts for accuracy:
Point 1 (y-intercept): $(I_1, V_1) = (0 \, \text{A}, 5.5 \, \text{V})$
Point 2 (x-intercept, estimated from the graph): $(I_2, V_2) = (4.5 \, \text{A}, 0 \, \text{V})$
Calculate the gradient ( $m$ ) of the line.
$m = \frac{\Delta V}{\Delta I} = \frac{V_2 - V_1}{I_2 - I_1}$ $m = \frac{0 \, \text{V} - 5.5 \, \text{V}}{4.5 \, \text{A} - 0 \, \text{A}}$ $m = \frac{-5.5 \, \text{V}}{4.5 \, \text{A}} \approx -1.222 \, \Omega$
Determine the internal resistance ( $r$ ).
$r = -m$ $r = -(-1.222 \, \Omega) = 1.222 \, \Omega$
The internal resistance of the battery is $\boxed{\text{1.22 \Omega}}$ .

8.2.1

To calculate the current in the $8 \, \Omega$ resistor, we use Ohm's law.
Apply Ohm's law ( $I = \frac{V}{R}$ ) to the $8 \, \Omega$ resistor.
$I_{8\Omega} = \frac{V_{8\Omega}}{R_{8\Omega}}$ $I_{8\Omega} = \frac{21.84 \, \text{V}}{8 \, \Omega}$ $I_{8\Omega} = 2.73 \, \text{A}$
The current in the $8 \, \Omega$ resistor is $\boxed{\text{2.73 A}}$ .

8.2.2

To calculate the equivalent resistance of the resistors in parallel, we use the formula for two parallel resistors.
Calculate the equivalent resistance for the $30 \, \Omega$ and $20 \, \Omega$ resistors in parallel.
$R_{parallel} = \frac{R_1 \times R_2}{R_1 + R_2}$ $R_{parallel} = \frac{30 \, \Omega \times 20 \, \Omega}{30 \, \Omega + 20 \, \Omega}$ $R_{parallel} = \frac{600 \, \Omega^2}{50 \, \Omega} = 12 \, \Omega$
The equivalent resistance of the resistors in parallel is $\boxed{\text{12 \Omega}}$ .

8.2.3

To determine the internal resistance $r$ of the battery, we use the total current and the EMF equation for a circuit with internal resistance.
Determine the total current ( $I_{total}$ ) in the circuit.
Since the $8 \, \Omega$ resistor is in series with the parallel combination, the current through it is the total current.
$I_{total} = I_{8\Omega} = 2.73 \, \text{A}$
Calculate the total external resistance ( $R_{ext}$ ).
This is the sum of the $8 \, \Omega$ resistor and the equivalent parallel resistance.
$R_{ext} = R_{8\Omega} + R_{parallel}$ $R_{ext} = 8 \, \Omega + 12 \, \Omega = 20 \, \Omega$
Use the EMF equation $\mathcal{E} = I_{total}(R_{ext} + r)$ and solve for $r$ .
Given $\mathcal{E} = 60 \, \text{V}$ .
$60 \, \text{V} = 2.73 \, \text{A} (20 \, \Omega + r)$
Divide both sides by $2.73 \, \text{A}$ :
$\frac{60 \, \text{V}}{2.73 \, \text{A}} = 20 \, \Omega + r$ $21.978... \, \Omega = 20 \, \Omega + r$ $r = 21.978... \, \Omega - 20 \, \Omega$ $r = 1.978... \, \Omega$
The internal resistance $r$ of the battery is $\boxed{\text{1.98 \Omega}}$ .

8.2.4

To calculate the heat dissipated in the external circuit, we use the formula $Q = I^2 R t$ .
Identify the values.
Total current ( $I_{total}$ ) = $2.73 \, \text{A}$ (from 8.2.1)
Total external resistance ( $R_{ext}$ ) = $20 \, \Omega$ (from 8.2.3)
Time ( $t$ ) = $0.2 \, \text{s}$
Calculate the heat dissipated ( $Q$ ).
$Q = I_{total}^2 R_{ext} t$ $Q = (2.73 \, \text{A})^2 \times 20 \, \Omega \times 0.2 \, \text{s}$ $Q = 7.4529 \, \text{A}^2 \times 4 \, \Omega \cdot \text{s}$ $Q = 29.8116 \, \text{J}$
The heat dissipated in the external circuit in 0.2 seconds is $\boxed{\text{29.81 J}}$ .

Explain Ohm's law and calculate the EMF and internal resistance of a battery from a V-I graph.

Step-by-step solution

8.1.1

8.1.2

8.1.3

8.1.4

Identify two points from the line of best fit.

Calculate the gradient ( $m$ ) of the line.

Determine the internal resistance ( $r$ ).

8.2.1

Apply Ohm's law ( $I = \frac{V}{R}$ ) to the $8 \, \Omega$ resistor.

8.2.2

Calculate the equivalent resistance for the $30 \, \Omega$ and $20 \, \Omega$ resistors in parallel.

8.2.3

Determine the total current ( $I_{total}$ ) in the circuit.

Calculate the total external resistance ( $R_{ext}$ ).

Use the EMF equation $\mathcal{E} = I_{total}(R_{ext} + r)$ and solve for $r$ .

8.2.4

Identify the values.

Calculate the heat dissipated ( $Q$ ).

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More Physics Questions

Explain Ohm's law and calculate the EMF and internal resistance of a battery from a V-I graph.

Step-by-step solution

8.1.1

8.1.2

8.1.3

8.1.4

Identify two points from the line of best fit.

Calculate the gradient (mmm) of the line.

Determine the internal resistance (rrr).

8.2.1

Apply Ohm's law (I=VRI = \frac{V}{R}I=RV​) to the 8 Ω8 \, \Omega8Ω resistor.

8.2.2

Calculate the equivalent resistance for the 30 Ω30 \, \Omega30Ω and 20 Ω20 \, \Omega20Ω resistors in parallel.

8.2.3

Determine the total current (ItotalI_{total}Itotal​) in the circuit.

Calculate the total external resistance (RextR_{ext}Rext​).

Use the EMF equation E=Itotal(Rext+r)\mathcal{E} = I_{total}(R_{ext} + r)E=Itotal​(Rext​+r) and solve for rrr.

8.2.4

Identify the values.

Calculate the heat dissipated (QQQ).

Still stuck on this one?

More Physics Questions

Calculate the gradient ( $m$ ) of the line.

Determine the internal resistance ( $r$ ).

Apply Ohm's law ( $I = \frac{V}{R}$ ) to the $8 \, \Omega$ resistor.

Calculate the equivalent resistance for the $30 \, \Omega$ and $20 \, \Omega$ resistors in parallel.

Determine the total current ( $I_{total}$ ) in the circuit.

Calculate the total external resistance ( $R_{ext}$ ).

Use the EMF equation $\mathcal{E} = I_{total}(R_{ext} + r)$ and solve for $r$ .

Calculate the heat dissipated ( $Q$ ).