Question

The circuit diagram shows three resistors of \(2 \Omega, 4 \Omega\) and \(R \Omega\) connected to a battery of e.m.f. 2 V and internal resistance \(3 \Omega\). A main current of 0.25 A flows through the circuit.

1. What is the p.d. across the 4 resistor?
2. Calculate the p.d. across the internal resistance of the cell.
3. What is the p.d. across the R or 22 resistor?
4. Calculate the value of R.

Answer 1

Given: EMF, \(E=2\,V\)

Internal resistance, \(r=3\,\Omega\)

Main current, \(I=0.25\,A\)

(a) Using Ohm’s law:

\[ V=IR \]

\[ =0.25\times4 \]

\[ {=1\,V} \]

(b) \[ V=Ir\]

\[=0.25\times3 \]

\[ {=0.75\,V} \]

(c) The 2 Ω and R Ω resistors are connected in parallel.

\[ V=E-(V_1+V_2) \]

\[ V=2-(1+0.75) \]

\[ V=2-1.75\]

\[ {V=0.25\,V} \]

(d) Current through the 2 Ω resistor:

\[ I_1=\frac{0.25}{2}\]

\[=0.125\,A \]

Current through R Ω:

\[ I_2=0.25-0.125\]

\[=0.125\,A \]

Using Ohm’s law:

\[ R=\frac{V}{I_2}\]

\[=\frac{0.25}{0.125} \]

\[ {=2\,\Omega} \]