39. A 6.0-V battery delivers a 0.50-A current to an electric motor connected across its terminals. What power is delivered to the motor? ) 40. A 50- resistance causes a current of 0.005A to flow through a circuit connected to a battery. What is the power in the circuit? uter memory will the word FRICT 41. A 100-W light bulb is connected to a 120-V electric line. What is the current that the light bulb draws? 0.8 al evew frigil nishes seview saeril to rilpneleysw ar

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### Electrical Power and Current Problems

**Problem 39:**
A 6.0-V battery delivers a 0.50-A current to an electric motor connected across its terminals. What power is delivered to the motor?

**Problem 40:**
A 50-Ω resistance causes a current of 0.005A to flow through a circuit connected to a battery. What is the power in the circuit?

**Problem 41:**
A 100-W light bulb is connected to a 120-V electric line. What is the current that the light bulb draws?

These problems are designed to help students understand basic electrical principles such as voltage, current, resistance, and power. Students can solve these problems using Ohm's Law and the power formula: 

\[ P = V \times I \]

Where:
- \( P \) is the power in watts (W),
- \( V \) is the voltage in volts (V),
- \( I \) is the current in amperes (A). 

Students may need to manipulate formulas and convert units for accurate solutions.
Transcribed Image Text:### Electrical Power and Current Problems **Problem 39:** A 6.0-V battery delivers a 0.50-A current to an electric motor connected across its terminals. What power is delivered to the motor? **Problem 40:** A 50-Ω resistance causes a current of 0.005A to flow through a circuit connected to a battery. What is the power in the circuit? **Problem 41:** A 100-W light bulb is connected to a 120-V electric line. What is the current that the light bulb draws? These problems are designed to help students understand basic electrical principles such as voltage, current, resistance, and power. Students can solve these problems using Ohm's Law and the power formula: \[ P = V \times I \] Where: - \( P \) is the power in watts (W), - \( V \) is the voltage in volts (V), - \( I \) is the current in amperes (A). Students may need to manipulate formulas and convert units for accurate solutions.
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