QUESTION IMAGE
Question
- calculate use ohms law to find the voltage needed to make a current of 3 a in a resistance of 9 ω. show your work
- calculate how much electrical energy, in kilowatt-hours, does a 200 w light bulb use if it is left on for 13 h? show your work
- apply concepts why does increasing the voltage of a wire have the same effect on the current that decreasing the resistance has?
Question 3
Step1: Recall Ohm's law formula
Ohm's law is \(V = IR\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance.
Step2: Substitute the given values
Given \(I = 3\space A\) and \(R=11\space\Omega\). Substitute into the formula: \(V=(3\space A)\times(11\space\Omega)\)
Step3: Calculate the voltage
\(V = 33\space V\)
Step1: Convert power to kilowatts
The power of the bulb \(P = 200\space W=0.2\space kW\) (since \(1\space kW = 1000\space W\)).
Step2: Use the energy formula \(E=Pt\)
The time \(t = 13\space h\). Substitute \(P = 0.2\space kW\) and \(t = 13\space h\) into \(E=Pt\). So \(E=(0.2\space kW)\times(13\space h)\)
Step3: Calculate the energy
\(E = 2.6\space kWh\)
According to Ohm's law \(I=\frac{V}{R}\). When \(V\) increases (assuming \(R\) is constant), \(I\) increases. When \(R\) decreases (assuming \(V\) is constant), \(I=\frac{V}{R}\) also increases. Mathematically, increasing \(V\) (numerator) and decreasing \(R\) (denominator) both lead to an increase in \(I\) based on the relationship \(I=\frac{V}{R}\)
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\(33\space V\)