Two resistances 2omega and 3omega are in parallel. The combination is in series with 15omega resistance and a power supply of voltage V. There is a current of 3omega resistance. What are the values of the current and votage across the power supply (a)3a and 10.5v(b) 4a and 9v (c)4a and 12v(d) 12a and 18v
R1 = 2 Megohms?
R2 = 3 megohms?
R3 = 15 Megohms?
I = 3 Microamps?
Please clarify your problem.
To find the values of the current and voltage across the power supply, we can analyze the given circuit step by step.
First, let's consider the combination of the two resistances, 2Ω and 3Ω, which are in parallel. When resistors are connected in parallel, their total resistance (R) can be calculated using the formula:
1/R_total = 1/R1 + 1/R2
In this case, substituting the values:
1/R_total = 1/2Ω + 1/3Ω
Simplifying:
1/R_total = 3/6Ω + 2/6Ω
1/R_total = 5/6Ω
Therefore, the total resistance (R_total) is 6/5Ω, or 1.2Ω.
Now, we have the series combination of the total resistance (1.2Ω) and a 15Ω resistor. The total resistance (R_total) in this series combination is the sum of the individual resistances:
R_total = 1.2Ω + 15Ω
R_total = 16.2Ω
Given that the current passing through the 3Ω resistor is 3A, we can use Ohm's Law (V = IR) to find the value of V, the voltage across the power supply.
V = I * R_total
V = 3A * 16.2Ω
V = 48.6V
So, the voltage across the power supply is 48.6V.
Also, since the current passing through the 3Ω resistor is 3A, this current is the same for the total circuit. Therefore, the current across the power supply is also 3A.
Hence, the correct answer is (a) 3A and 48.6V.