A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a

resistor. The circuit does not contain a voltmeter or an ammeter. When the circuit is operating, the

total current through the circuit is 0.50 ampere.

Determine the equivalent resistance of the circuit.

Considering the unknown resistence as R and using the Ohm's First Law, we have:

\(i= \frac{V}{R_{eq}} \\ 0.5= \frac{12}{R+10} \\ R+10=24 \\ R=14-Ohm\)

The equivalent resistence is given by the resistor series with the lamp resistence.

\(R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}\)

If you notice any mistake in my english, please let me know, because i am not native.

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A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and aresistor. The circuit does not contain a voltmeter or an ammeter. When the circuit is operating, thetotal current through the circuit is 0.50 ampere.Determine the equivalent resistance of the circuit.

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