Question

but 9 volt battery is connected to a 4 ohm resistor and the 5 Ohm resistor as shown in the diagram. How much current flows through the 4 ohm resistor ?

Answer 1

Given data

The resistance of the first resistor is R1 = 4 ohm

The resistance of the second resistor is R2 = 5 ohm

The potential difference of the battery is V = 9 V

The resistors are connected in series. The expression for the equivalent resistance is given as:

`\begin{gathered} R=R_1+R_2_{} \\ R=4\text{ }\Omega+5\text{ }\Omega \\ R=9\Omega \end{gathered}`

The expression for the current in the 4-ohm resistor is given as:

`\begin{gathered} I=(V)/(R) \\ I=\frac{9\text{ V}}{9\text{ }\Omega} \\ I=1\text{ A} \end{gathered}`

Thus, the magnitude of the current flows through the 4-ohm resistor is 1 A.