Answer:
The current of 10 mA is flowing through the 250-Ω resistor.
Explanation:
The values of four resistors 150 ohms, 250 ohms, 350 ohms and 450 ohms are connected in series with a 12 V battery. The equivalent resistance in case of series combination is given by :
[tex]R_{eq}=150+250+350+450[/tex]
[tex]R_{eq}=1200\ \Omega[/tex]
In series combination, the current flowing throughout the resistors is same. It is given by using Ohm's law as :
[tex]V=IR_{eq}[/tex]
[tex]I=\dfrac{V}{R_{eq}}[/tex]
[tex]I=\dfrac{12}{1200}[/tex]
[tex]I=0.01\ A[/tex]
or
[tex]I=10\ mA[/tex]
So, the current of 10 mA is flowing through the 250-Ω resistor. Hence, this is the required solution.
