Four resistors (R1 = 150 Ω, R2 = 250 Ω, R3 = 350 Ω, R4 = 450 Ω) are connected in series with a 12-V battery. What is the current that flows through the 250-Ω resistor?

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 :

In series combination, the current flowing throughout the resistors is same. It is given by using Ohm's law as :

or

So, the current of 10 mA is flowing through the 250-Ω resistor. Hence, this is the required solution.

f=ma newton's second law 7x8=56n