Question

A circuit with a 22 V battery a 15 Ω resistor and 5 Ω resistor in series. What is the current throughout the circuit?

Answer 1

Given data:

* The voltage across the battery is V = 22 volts.

* The resistance of resistors connected in series is,

`\begin{gathered} R_1=15\text{ ohm} \\ R_2=5\text{ ohm} \end{gathered}`

Solution:

As the resistors are connected in series, thus, the equivalent resistance of the circuit is,

`\begin{gathered} R_(eq)=R_1+R_2 \\ R_(eq)=15+5 \\ R_(eq)=20\text{ ohm} \end{gathered}`

According to Ohm's law, the current through the circuit is,

`\begin{gathered} V=IR_(eq) \\ I=(V)/(R_(eq)) \end{gathered}`

Substituting the known values,

`\begin{gathered} I=(22)/(20) \\ I=1.1\text{ A} \end{gathered}`

Thus, the current through the circuit is 1.1 A.