Equivalent resistance for **two** resistors in parallel:

$\overline{){{\mathbf{R}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}\frac{{\mathbf{R}}_{\mathbf{1}}{\mathbf{R}}_{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{R}}_{\mathbf{2}}}}$

Current when the switch is open:

${\mathit{i}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{\epsilon}}{\mathbf{r}\mathbf{+}{\mathbf{R}}_{\mathbf{5}\mathbf{.}\mathbf{0}}}$

1.636 = ε/(r + 5.0)

When the switch is closed, the current through the 5.0-Ω resistor is:

i_{1} = 1.565A

Voltage, V = i_{1}R

V = i_{1}(5.0) = (1.565)(5.0) = 7.825 V

The potential across the 10.0-Ω resistor is:

V = 7.825 V

Current across the 10.0-Ω resistor is:

i_{2} = 7.825/10.0 = 0.7825A

When teh switch is closed,

i = i_{1} + i_{2} = 1.565 + 0.7825 = 2.3475A

The resistors are in parallel:

R_{eq} = (5.0)(10.0)/(5.0 + 10.0) = 3.333Ω

A) What is the emf of the battery in the figure ?

B) What is the internal resistance of the battery in the figure?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Combining Resistors in Series & Parallel concept. You can view video lessons to learn Combining Resistors in Series & Parallel. Or if you need more Combining Resistors in Series & Parallel practice, you can also practice Combining Resistors in Series & Parallel practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Arnold's class at PURDUE.