Question

A battery with V = 2.51 V is connected to three resistors as shown in the figure, where R₁ = 2.13 ?, R₂ = 4.63 ?, and R₃ = 5.25 ?.

a) Find the potential drop across resistor 3.
1.334 V

b) Find the current in resistor 1.

1.178 A

The correct answers are in bold but I would like someone to show my how you get them.

Answer 1

(a) we know that, resistance R₂ & R₃ connected in series.

R_eq = (4.63 $\Omega$ ) + (5.25 $\Omega$ )

R_eq = 9.88 $\Omega$

current flowing in the 2nd loop,   I = V / R_eq $\Rightarrow$ (2.51 V) / (9.88 $\Omega$ )

I = 0.254 A

Now, the potential drop across resistor 3 will be given as :

V₃ = I R₃ $\Rightarrow$ (0.254 A) (5.25 $\Omega$ )

V₃ = 1.334 V

(b) The current in resistor 1 will be given as :

using an ohm's law,    V = I R₁

where, V = voltage of the battery = 2.51 V

R₁ = resistance = 2.13 $\Omega$

then, we have

(2.51 V) = I (2.13 $\Omega$ )

I = 1.178 A