Just to be clear: the resistors in segment A-B are 8 ohm and 3 ohm, while those in segment CD are 12 ohm and 6 ohm.
We can use the concepts of series and parallel circuits to solve the problem. Let us assume that the current is I1 and I2 in segments A-B and C-D, respectively. For the circuit involving battery and segment A-B, we can write,
-4 + 8 I1 + 3 I1 =...
Just to be clear: the resistors in segment A-B are 8 ohm and 3 ohm, while those in segment CD are 12 ohm and 6 ohm.
We can use the concepts of series and parallel circuits to solve the problem. Let us assume that the current is I1 and I2 in segments A-B and C-D, respectively. For the circuit involving battery and segment A-B, we can write,
-4 + 8 I1 + 3 I1 = 0
Solving this equation, we get, I1 = 4/11 A
Similarly, for circuit involving battery and segment C-D, we can write,
-4 + 12 I2 + 6 I2 = 0
Solving this equation, we get I2 = 4/18 A
In a series circuit (such as in the segment C-D), the same current flows through each resistor, thus the current flowing through the 6 ohm resistor is 4/18 A or 0.22 A.
The potential difference across the 12 ohm resistor can be calculated as the product of current and resistance, that is
Potential difference = 12 I2 = 2.67 V.
No comments:
Post a Comment