B.5 Current and circuits

Solution outline

1. $I = \dfrac{\Delta Q}{\Delta t} \Rightarrow \Delta Q = I\,\Delta t.$
2. $\Delta t = 5.0\ \text{min} = 300\ \mathrm{s}.$
3. $\Delta Q = 0.80\times300 = 240\ \mathrm{C}.$
4. $\text{Current is defined as the rate of flow of charge through a cross‑section of the conductor.}$

Solution outline

1. $R = \rho\,\dfrac{L}{A} = 1.7\times10^{-8}\times\dfrac{3.0}{1.5\times10^{-6}}.$
2. $R = 1.7\times10^{-8}\times2.0\times10^{6} = 0.034\ \Omega.$
3. $\text{Increasing length increases resistance proportionally; increasing cross‑sectional area reduces resistance because there is more room for charge carriers to move.}$

Solution outline

1. $\text{An ohmic resistor has a straight‑line V–I graph through the origin; R = V/I is constant.}$
2. $\text{A filament lamp has a curve that becomes less steep at higher currents; the resistance increases as the filament heats up.}$
3. $\text{In the lamp, higher current raises temperature, which increases lattice vibrations and makes it harder for electrons to drift, so V is no longer proportional to I.}$

Solution outline

1. $\text{Series: } R_{s} = 4.0 + 6.0 = 10.0\ \Omega.$
2. $I_{s} = \dfrac{V}{R_{s}} = \dfrac{12}{10.0} = 1.2\ \mathrm{A}.$
3. $\text{Parallel: } \dfrac{1}{R_{p}} = \dfrac{1}{4.0} + \dfrac{1}{6.0} = 0.25 + 0.167 = 0.417.$
4. $R_{p} \approx 2.4\ \Omega,\quad I_{p} = \dfrac{12}{2.4} = 5.0\ \mathrm{A}.$
5. $\text{Series circuits have the same current everywhere and voltages that add; parallel circuits have the same voltage across each branch and currents that add.}$

Solution outline

1. $\text{Total resistance } R_{\text{tot}} = R + r = 2.6 + 0.40 = 3.0\ \Omega.$
2. $I = \dfrac{\varepsilon}{R_{\text{tot}}} = \dfrac{1.5}{3.0} = 0.50\ \mathrm{A}.$
3. $V = \varepsilon - Ir = 1.5 - 0.50\times0.40 = 1.3\ \mathrm{V}.$
4. $\text{The emf is the energy supplied per coulomb; the internal resistance causes a drop in terminal voltage when current flows.}$

Solution outline

1. $P = IV = 4.0\times12 = 48\ \mathrm{W}.$
2. $\Delta t = 10\ \text{min} = 600\ \mathrm{s},\quad E = P\,\Delta t = 48\times600 = 2.9\times10^{4}\ \mathrm{J}.$
3. $\text{In a heater this energy transfer to thermal energy is useful; in ordinary circuit components it represents unwanted power loss and heating.}$

Solution outline

1. $\text{A variable resistor changes the total series resistance in the circuit.}$
2. $\text{Increasing its resistance reduces the current } I = V/R_{\text{total}} \text{ supplied to the lamp.}$
3. $\text{Lower current means lower power } P = I^{2}R_{\text{lamp}}, \text{ so the lamp becomes dimmer.}$