Question

Two resistors of equal resistance are connected in series with each other and are connected to a battery that produces a potential difference of 8 V. If the current is 0.2 A what is the value of each resistance?

Answer 1

Question

Given

$$\begin{array}{l} V = 8 \mathrm{~V} \\ I = 0.2 \mathrm{~A} \\ R_1 = R_2 = R \\ \end{array}$$

Need to find: resistance $R$.

Solution:

If two resistors connected in series then their common resistance equal to the sum of their resistance. So, we have that the resistance of the circuit is $R_c = R_1 + R_2 = R + R = 2R$. So, as we know that the voltage is $V = 8\mathrm{~V}$ and the current is $I = 0.2\mathrm{~A}$ we can find the resistance:

$$I = \frac{V}{R_c} = \frac{V}{2R} \Rightarrow V = 2IR \Rightarrow R = \frac{V}{2I} = \frac{8}{2 \cdot 0.2} = 20 \mathrm{~Ohm}.$$

So, the value of each resistance is $20\mathrm{Ohm}$.

Answer: 20 Ohm.