Answer :
We are given two problems.
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Problem 5: A simple circuit with a 9 V battery and a resistance of $150\,\Omega$ has a charge of $1\,\text{C}$ passing through it. We want to determine the time taken for the charge to pass.
**Step 1. Calculate the current using Ohm’s law.**
Ohm’s law states that
$$
I = \frac{V}{R},
$$
where:
- $V = 9\,\text{V}$ is the battery voltage,
- $R = 150\,\Omega$ is the resistance.
Thus, the current is
$$
I = \frac{9}{150} = 0.06\,\text{A}.
$$
**Step 2. Calculate the time using the relationship between charge and current.**
The relationship is given by:
$$
Q = I \times t,
$$
so the time $t$ is
$$
t = \frac{Q}{I}.
$$
Here, $Q = 1\,\text{C}$ is the charge that passes through. Therefore,
$$
t = \frac{1}{0.06} \approx 16.67\,\text{s}.
$$
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Problem 6: An electric stove element creates $150\,\text{J}$ of heat at $240\,\text{V}$ over a $5$ minute period. We need to determine the current in the stove element.
**Step 1. Convert the time from minutes to seconds.**
Since $1$ minute is $60$ seconds:
$$
5\,\text{minutes} \times 60\,\frac{\text{seconds}}{\text{minute}} = 300\,\text{s}.
$$
**Step 2. Use the energy relation for electrical devices.**
The energy consumed is related to power and time by:
$$
E = P \times t,
$$
and the power is also expressed by:
$$
P = I \times V.
$$
Combining these gives:
$$
E = I \times V \times t.
$$
Solving for the current $I$, we have:
$$
I = \frac{E}{V \times t}.
$$
Substitute the known values:
- $E = 150\,\text{J}$,
- $V = 240\,\text{V}$,
- $t = 300\,\text{s}$,
so that
$$
I = \frac{150}{240 \times 300} \approx 0.002083\,\text{A}.
$$
–––––––––––––––––––––––––––
**Final Answers:**
1. In the simple circuit, the current is $0.06\,\text{A}$ and it takes approximately $16.67$ seconds for $1\,\text{C}$ of charge to pass through.
2. For the electric stove element, the current is approximately $0.002083\,\text{A}$.
–––––––––––––––––––––––––––
Problem 5: A simple circuit with a 9 V battery and a resistance of $150\,\Omega$ has a charge of $1\,\text{C}$ passing through it. We want to determine the time taken for the charge to pass.
**Step 1. Calculate the current using Ohm’s law.**
Ohm’s law states that
$$
I = \frac{V}{R},
$$
where:
- $V = 9\,\text{V}$ is the battery voltage,
- $R = 150\,\Omega$ is the resistance.
Thus, the current is
$$
I = \frac{9}{150} = 0.06\,\text{A}.
$$
**Step 2. Calculate the time using the relationship between charge and current.**
The relationship is given by:
$$
Q = I \times t,
$$
so the time $t$ is
$$
t = \frac{Q}{I}.
$$
Here, $Q = 1\,\text{C}$ is the charge that passes through. Therefore,
$$
t = \frac{1}{0.06} \approx 16.67\,\text{s}.
$$
–––––––––––––––––––––––––––
Problem 6: An electric stove element creates $150\,\text{J}$ of heat at $240\,\text{V}$ over a $5$ minute period. We need to determine the current in the stove element.
**Step 1. Convert the time from minutes to seconds.**
Since $1$ minute is $60$ seconds:
$$
5\,\text{minutes} \times 60\,\frac{\text{seconds}}{\text{minute}} = 300\,\text{s}.
$$
**Step 2. Use the energy relation for electrical devices.**
The energy consumed is related to power and time by:
$$
E = P \times t,
$$
and the power is also expressed by:
$$
P = I \times V.
$$
Combining these gives:
$$
E = I \times V \times t.
$$
Solving for the current $I$, we have:
$$
I = \frac{E}{V \times t}.
$$
Substitute the known values:
- $E = 150\,\text{J}$,
- $V = 240\,\text{V}$,
- $t = 300\,\text{s}$,
so that
$$
I = \frac{150}{240 \times 300} \approx 0.002083\,\text{A}.
$$
–––––––––––––––––––––––––––
**Final Answers:**
1. In the simple circuit, the current is $0.06\,\text{A}$ and it takes approximately $16.67$ seconds for $1\,\text{C}$ of charge to pass through.
2. For the electric stove element, the current is approximately $0.002083\,\text{A}$.
