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19 December, 08:48

answers A 17.0 resistor and a 6.0 resistor are connected in series with a battery. The potential difference across the 6.0 resistor is measured as 15 V. Find the potential difference across the battery.

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Answers (2)
  1. 19 December, 10:45
    0
    57.5volts

    Explanation

    Since the resistors given are connected in series, therefore the same current will flows in both resistors but different voltage will be across them.

    If the potential difference across the 6.0 resistor is measured as 15 V, then the current in the 6.0 resistor can be gotten using ohm's law formula V = IR where;

    V is the potential difference = 15V

    I is the current = ?

    R is the resistance = 6.0

    I = V/R

    I = 15/6

    I = 2.5A

    This means a current of 2.5A flows in the 6.0 resistor.

    This same current will therefore flow in the 17.0 resistors and also be the total current flowing in the circuit.

    To calculate the potential difference E across the battery, we will use the expression E = IRt where;

    I is the total current flowing in the circuit = 2.5A

    Rt is the effective resistance = 17+6 = 23.0ohms (since they are connected in series)

    E = 2.5*23

    E = 57.5volts

    The potential difference across the battery is 57.5volts
  2. 19 December, 12:15
    0
    Potential difference across the battery = 57.5 V

    Explanation:

    We can solve this problem more efficiently if we use voltage divider rule to find out the voltage across the battery.

    Voltage divider rule to find voltage across 6 Ω resistor

    V₆ = V (R₁₇ / (R₁₇+R₆))

    Where V₆ is the voltage across 6 Ω resistor and V is the overall voltage across battery, re-arrange the equation to find V

    V = V₆ (R₁₇ + R₆) / R₁₇))

    V = 15 ((17 + 6) / 17)

    V = 15 (23/6)

    V = 57.5 V

    Bonus:

    The total voltage in a circuit is the sum of voltage drops at each resistor so we can find out the voltage at 17 Ω resistor

    V = V₆ + V₁₇

    V₁₇ = V - V₆

    V₁₇ = 57.5 - 15

    V₁₇ = 42.5 V
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