Current, potential difference and resistance

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Question 8
Easy

A student built a simple circuit consisting of a battery, an ammeter, and a lamp connected in series.

A circuit diagram with a single loop containing a DC power source (battery) labeled 5.0 V at the top, an ammeter on the left wire, and a single lamp (represented by a circle with an X inside) on the bottom wire.

a.

In 25 seconds25\text{ seconds}25 seconds, 20 coulombs20\text{ coulombs}20 coulombs of charge flow through the battery.

Calculate the current in the battery.

Use the equation: current=charge flowtime\text{current} = \frac{\text{charge flow}}{\text{time}}current=timecharge flow​

[2]
b.

There is a potential difference of 5.0 V5.0\text{ V}5.0 V across the battery.

Calculate the energy transferred by the battery when 70 coulombs70\text{ coulombs}70 coulombs of charge flows through it.

Use the equation: energy transferred=charge flow×potential difference\text{energy transferred} = \text{charge flow} \times \text{potential difference}energy transferred=charge flow×potential difference

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c.

Write down the equation that links current (III), potential difference (VVV), and resistance (RRR).

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d.

The potential difference across the lamp is 5.0 V5.0\text{ V}5.0 V and the current through it is 0.80 A0.80\text{ A}0.80 A.

Calculate the resistance of the lamp.

[2]

Current, potential difference and resistance Questions

  1. GCSE
  2. /Physics
  3. /Current, potential difference and resistance