A circuit has three resistors connected in series. Resistor R2 has a resistance of 220 Ω and a voltage drop of 44 V. What is the current flow through resistor R3?
The value of the resistance in the series circuit is \(\mathrm{R}_{2}=220 \Omega\).
In the series circuit, the voltage drop across the resistor \(R_{2}\) is \(\mathrm{E}_{2}=44 \mathrm{~V}\).
In the series circuit, the current flowing is equal in all the elements of the circuit,
From Ohm's law to the resistor \(R_{2}\), the current \(I\) flowing across it is given by
\(\mathrm{I}=\frac{\mathrm{E}_{2}}{\mathrm{R}_{2}}\)
\(=\frac{44}{220}\)
\(=0.2 \mathrm{~A}\)
Hence, the current flowing through the resistor \(\mathrm{R}_{2}\) is
\(\mathrm{I}=0.2 \mathrm{~A}\).