12 exam-style questions on current, resistance, circuits, potential dividers and internal resistance. Tap Show answer for full working.
A current of \(2.5\,\mathrm{A}\) flows through a resistor for \(4.0\) minutes. Calculate the charge that passes and the number of electrons that flow past a point in the resistor. (\(e = 1.6\times10^{-19}\,\mathrm{C}\).)
\(Q = It = 2.5\times240 = 600\,\mathrm{C}\).
Number of electrons \(= Q/e = 600/1.6\times10^{-19} = 3.75\times10^{21}\).
A \(12\,\mathrm{V}\) battery is connected to a \(6.0\,\Omega\) resistor. Calculate the current in the circuit and the power dissipated in the resistor.
\(I = V/R = 12/6 = 2.0\,\mathrm{A}\).
\(P = I^{2}R = 4\times6 = 24\,\mathrm{W}\) (or \(V^{2}/R = 24\,\mathrm{W}\)).
Two resistors of \(4.0\,\Omega\) and \(6.0\,\Omega\) are connected in series across a \(20\,\mathrm{V}\) supply. Calculate the total resistance, the current, and the potential difference across the \(6.0\,\Omega\) resistor.
\(R_{\mathrm{tot}} = 10\,\Omega\); \(I = 20/10 = 2.0\,\mathrm{A}\).
\(V_{6} = IR = 2\times6 = 12\,\mathrm{V}\).
The same two resistors (\(4.0\,\Omega\) and \(6.0\,\Omega\)) are now connected in parallel across the \(20\,\mathrm{V}\) supply. Calculate the total resistance and the current through the \(4.0\,\Omega\) resistor.
\(1/R = 1/4 + 1/6 = 5/12 \Rightarrow R = 2.4\,\Omega\).
\(I_{4} = V/R = 20/4 = 5.0\,\mathrm{A}\).
A cell has EMF \(1.5\,\mathrm{V}\) and internal resistance \(0.50\,\Omega\). It is connected to a \(2.5\,\Omega\) resistor. Calculate the current in the circuit and the terminal potential difference.
\(I = \mathcal{E}/(R+r) = 1.5/(2.5+0.5) = 0.50\,\mathrm{A}\).
\(V = IR = 0.50\times2.5 = 1.25\,\mathrm{V}\) (or \(\mathcal{E}-Ir = 1.25\,\mathrm{V}\)).
A potential divider consists of a \(3.0\,\mathrm{k}\Omega\) resistor and a \(5.0\,\mathrm{k}\Omega\) resistor in series across a \(12\,\mathrm{V}\) supply. Calculate the output voltage across the \(5.0\,\mathrm{k}\Omega\) resistor.
\(V_{\mathrm{out}} = V\times R_{2}/(R_{1}+R_{2}) = 12\times5/(3+5) = 7.5\,\mathrm{V}\).
A wire of resistivity \(1.7\times10^{-8}\,\Omega\,\mathrm{m}\), length \(2.0\,\mathrm{m}\) and diameter \(0.40\,\mathrm{mm}\) carries a current of \(1.5\,\mathrm{A}\). Calculate the resistance of the wire and the potential difference across it.
\(A = \pi(0.20\times10^{-3})^{2} = 1.26\times10^{-7}\,\mathrm{m}^{2}\).
\(R = \rho L/A = 1.7\times10^{-8}\times2.0/1.26\times10^{-7} = 0.270\,\Omega\).
\(V = IR = 1.5\times0.270 = 0.405\,\mathrm{V}\).
Explain what is meant by the EMF of a cell and its internal resistance. A cell of EMF \(3.0\,\mathrm{V}\) and internal resistance \(1.0\,\Omega\) is connected to a variable resistor \(R\). Sketch a graph of terminal PD against current and explain how the EMF and internal resistance can be determined from the graph.
EMF: the energy transferred per unit charge by the cell when charge flows round a complete circuit (or open-circuit terminal PD). Internal resistance: the resistance of the materials inside the cell that dissipates energy as heat.
Graph of \(V\) vs \(I\) is a straight line with intercept \(\mathcal{E}\) on the \(V\)-axis and gradient \(-r\). Short-circuit current intercept on \(I\)-axis is \(\mathcal{E}/r\).
A battery of EMF \(12\,\mathrm{V}\) and internal resistance \(0.80\,\Omega\) is connected to two resistors of \(3.0\,\Omega\) and \(6.0\,\Omega\) in parallel. Calculate the current drawn from the battery and the power dissipated in the \(3.0\,\Omega\) resistor.
Parallel combination: \(R_p = (3\times6)/(3+6) = 2.0\,\Omega\).
Total resistance \(= 2.0 + 0.80 = 2.8\,\Omega\); \(I_{\mathrm{total}} = 12/2.8 = 4.29\,\mathrm{A}\).
Terminal PD \(= I\times R_p = 4.29\times2.0 = 8.57\,\mathrm{V}\).
\(I_{3} = 8.57/3 = 2.86\,\mathrm{A}\); \(P_{3} = I^{2}R = (2.86)^{2}\times3 = 24.5\,\mathrm{W}\).
A thermistor has resistance \(4.0\,\mathrm{k}\Omega\) at \(20^\circ\mathrm{C}\) and \(1.5\,\mathrm{k}\Omega\) at \(60^\circ\mathrm{C}\). It is connected in series with a fixed \(2.0\,\mathrm{k}\Omega\) resistor across a \(9.0\,\mathrm{V}\) supply to form a potential divider, with the output taken across the fixed resistor. Calculate the output voltage at each temperature and state whether the output rises or falls as temperature increases.
At \(20^\circ\mathrm{C}\): \(V_{\mathrm{out}} = 9\times2/(4+2) = 3.0\,\mathrm{V}\).
At \(60^\circ\mathrm{C}\): \(V_{\mathrm{out}} = 9\times2/(1.5+2) = 5.14\,\mathrm{V}\).
Output rises as temperature increases (thermistor resistance falls, larger share of voltage across fixed resistor).
In a circuit a cell of EMF \(\mathcal{E}\) and internal resistance \(r\) is connected to a load resistor \(R\). Show that the power delivered to the load is \(P = \mathcal{E}^{2}R/(R+r)^{2}\). Hence find the value of \(R\) for which \(P\) is a maximum, and state the maximum power in terms of \(\mathcal{E}\) and \(r\).
\(I = \mathcal{E}/(R+r)\); \(P = I^{2}R = \mathcal{E}^{2}R/(R+r)^{2}\).
To maximise, differentiate: let \(u=R\), \(v=(R+r)^{2}\); \(\mathrm{d}P/\mathrm{d}R = 0\) when \(v - u\cdot2(R+r) = 0 \Rightarrow (R+r)^{2} - 2R(R+r) = 0 \Rightarrow R+r - 2R = 0 \Rightarrow R = r\).
Maximum power \(= \mathcal{E}^{2}r/(2r)^{2} = \mathcal{E}^{2}/(4r)\).
A copper wire of length \(L\) and diameter \(d\) has resistance \(R\). A second wire of the same material has length \(2L\) and diameter \(2d\). Calculate the ratio of the resistance of the second wire to the first.
\(R = \rho L/A\); \(A \propto d^{2}\).
\(R_{2}/R_{1} = (L_{2}/L_{1})\times(A_{1}/A_{2}) = 2\times(d^{2}/(2d)^{2}) = 2\times(1/4) = 1/2\).