# Finding resistance for parallel resistors: how is my reasoning incorrect, and what is the correct reasoning?

I have the following circuit:

The voltmeter $$\V_1\$$ measures the voltage drop across the $$\9.00 \ \Omega\$$ resistor and has a reading of $$\4.50 \ \text{V}\$$.

The reading on ammeter $$\A_1\$$ is $$\I_1 = 0.50 \ \text{A}\$$, and the reading on ammeter $$\A_2\$$ is $$\I_2 = 0.38 \ \text{A}\$$.

I am now trying to determine the unknown resistance $$\R\$$. It seems to me that we have two resistors in parallel here, so we can use the formula $$\R_{\text{total}} = \dfrac{R_1 R_2}{ R_1 + R_2 }\$$. So we have that $$\R_{\text{total}} = \dfrac{R \times 4.0 \ \Omega}{ R + 4.0 \ \Omega }\$$. We can now incorporate this into Ohm's law, where $$\I = 0.38 \ \text{A}\$$ and $$\V = 6.0 \ \text{V} - 4.5 \ \text{V} = 1.5 \ \text{V}\$$, so that we get $$\1.5 = 0.38 \left( \dfrac{R \times 4.0 \ \Omega}{ R + 4.0 \ \Omega } \right) \ \Rightarrow R = 300 \ \Omega\$$. But I am told that the solution is actually $$\ I = 0.5 - 0.375 = 0.125 \ \text{A} \$$ (where I round $$\0.375\$$ to $$\0.38\$$) and $$\R = \dfrac{1.5}{0.125} = 12 \ \Omega\$$. How is my reasoning incorrect, and what is the correct reasoning that leads to $$\ I = 0.5 - 0.375 = 0.125 \ \text{A} \$$ and $$\R = \dfrac{1.5}{0.125} = 12 \ \Omega\$$?

• I2 is not a total current (I2 flows only via 4ohms resistor ). Thus the I_total = 0.5A So R_total = 1.5V/0.5A = 3Ohms
– G36
Commented May 12, 2022 at 12:25
• Does this answer your question? Calculating the reading on ammeter $A_1$: why am I incorrect?
– Bart
Commented May 12, 2022 at 13:01