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What is the terminal voltage if it is connected in series to a circuit with a total resistance of 92.9 Ω ?
A 6.00 V battery has an internal resistance of 0.803Ω . What is the terminal voltage if it is connected in series to a circuit with a total resistance of 92.9Ω?
5.99 V
5.87 V
5.91 V
5.89 V
3 Answers
- AshLv 74 years ago
terminal voltage = (circuit resistance/ total resistance) x battery voltage = [92.9/(92.9+0.803)]x6 = 5.95 V
- The GnosticLv 74 years ago
The total resistance in the circuit is 92.9 + 0.803 = 93.703 Ω.
Against 6 volts, that is approximately 0.064 amperes.
That amperage × 0.803 Ω ≈ 0.0514 volts lost to the internal resistance.
The terminal voltage is 5.9486 V, rounded.
- Anonymous4 years ago
No
