🔌 بنك الأسئلة: الدوائر الكهربائية البسيطة

📌 From the figure: \(R_1=8\,\Omega\) and \(R_3=6\,\Omega\), and \(R_{eq}=25\,\Omega\).

📌 In series: \(R_{eq}=R_1+R_2+R_3\Rightarrow25=8+R_2+6\).

\(R_2=25-14=11\,\Omega\).

📌 For identical resistors in series: \(R_{eq}=n\,R\) where \(n\) is the number of resistors.

\(120=5R\Rightarrow R=\dfrac{120}{5}=24\,\Omega\).

📌 In parallel: \(\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\).

\(\dfrac{1}{6}=\dfrac{1}{60}+\dfrac{1}{R_2}+\dfrac{1}{10}\).

\(\dfrac{1}{R_2}=\dfrac{10-1-6}{60}=\dfrac{3}{60}=\dfrac{1}{20}\Rightarrow R_2=20\,\Omega\).

📌 For identical resistors in parallel: \(R_{eq}=\dfrac{R}{n}\).

\(12=\dfrac{R}{5}\Rightarrow R=12\times5=60\,\Omega\).

📌 Series features: the equivalent resistance = the sum (larger than any one), the current is the same everywhere and the voltage across each resistor is proportional to its resistance.

📌 In series a break stops the whole circuit; the rest staying on is a parallel feature.

📌 In parallel the voltage is the same on all, so the current \(I=V/R\) is inversely proportional to the resistance: the larger resistor carries the smaller current.

📌 Voltage divider in series: \(V_1=\dfrac{R_1}{R_1+R_2}\,V_{total}\).

\(V_1=\dfrac{60}{100}\times5=3\,V\).

\(R_{eq}=\dfrac{4\times4}{4+4}=2\,\Omega\).

📌 The ammeter measures the total current: \(I=\dfrac{V}{R_{eq}}=\dfrac{6}{2}=3\,A\).

📌 The voltmeter has a very high resistance and is in series, so it practically blocks the current: \(I\approx0\).

📌 With zero current there is no voltage drop across the other elements, so the voltmeter reads the whole source voltage \(120\,V\).

📌 In series (figure A) the resistance is largest, while in parallel the equivalent resistance is smaller than any single one.

📌 Figure D shows all four resistors in parallel, which gives the smallest equivalent resistance.

📌 In series: \(R_1+R_2=9\) ... (1).

📌 In parallel: \(\dfrac{R_1R_2}{R_1+R_2}=2\Rightarrow R_1R_2=2\times9=18\) ... (2).

📌 Two numbers with sum \(9\) and product \(18\): \(3\) and \(6\); since \(R_2\gt R_1\), \(R_1=3\,\Omega,\ R_2=6\,\Omega\).

📌 The second and third in parallel: \(R_{23}=\dfrac{R}{2}\).

\(R_{eq}=R+\dfrac{R}{2}+R=\dfrac{5R}{2}\).

\(\dfrac{5R}{2}=5\Rightarrow R=2\,\Omega\).

📌 The branch with \(R_4\) is connected directly across the battery: \(V_4=6\,V\).

\(I_4=\dfrac{V_4}{R_4}=\dfrac{6}{4}=1.5\,A\).

📌 \(R_1\) and \(R_3\) in parallel: \(R_{13}=\dfrac{6\times3}{6+3}=2\,\Omega\).

📌 The result is in series with \(R_2\) (the battery is in its branch): \(R_{eq}=2+18=20\,\Omega\).

📌 The path \(A\to B\to C\) has three resistors in series \(5+5+5=15\,\Omega\) (two between \(A\) and \(B\) then one between \(B\) and \(C\)), the other path is identical, and the battery gives \(2\,V\) between \(A\) and \(C\).

📌 Voltage divider: \(V_{AB}=\dfrac{10}{15}\times2=\dfrac{4}{3}\,V\).

📌 Before closing: \(I_1=\dfrac{V}{R}\).

📌 After closing the two are in parallel \(R_{eq}=\dfrac{R}{2}\) so \(I_2=\dfrac{V}{R/2}=\dfrac{2V}{R}=2I_1\).

📌 \(R_{12}=\dfrac{8\times2}{8+2}=1.6\,\Omega\).

📌 The voltage across the two branches \(V=I\,R_{12}=2\times1.6=3.2\,V\).

\(I_1=\dfrac{V}{R_1}=\dfrac{3.2}{8}=0.4\,A\).

📌 Kirchhoff’s first law: the sum of the currents entering a junction = the sum of the currents leaving it.

\(I_1+I_2=I_3+I_4\Rightarrow4+I_2=6+5\Rightarrow I_2=7\,A\).

📌 All three in series \(300\,\Omega\); all three in parallel \(33.3\,\Omega\); two in parallel in series with the third \(100+50=150\,\Omega\); two in series in parallel with the third \(\dfrac{200\times100}{300}=66.7\,\Omega\).

📌 The smallest possible value is \(33.3\,\Omega\), so \(20\,\Omega\) is below it and cannot be obtained.

📌 A voltmeter has a very large resistance and is connected in parallel; connected in series it practically blocks the current.

📌 The ammeter has a very small resistance and is connected in series; the fuse and the rheostat are also connected in series without blocking the current.

📌 In series the current is the same in all and the power \(P=I^{2}R\) is directly proportional to the resistance.

📌 The largest resistance is \(R_3=8\,\Omega\), so its lamp is the brightest.

📌 The total current is the sum of the currents of the parallel branches: circuit A \(4+8+3=15\,A\) (exactly the fuse limit, no safety margin), B \(6+7+5=18\,A\), D \(2+10+4=16\,A\).

📌 Only circuit C gives \(4+3+6=13\,A\lt 15\,A\), so the fuse suits it.

\(R_{eq}=\dfrac{V}{I}=\dfrac{18}{0.6}=30\,\Omega\).

\(30=5+10+R_3\Rightarrow R_3=15\,\Omega\).

\(R_{total}=\dfrac{V}{I}=\dfrac{12}{0.5}=24\,\Omega\).

\(R_{added}=24-2=22\,\Omega\).

📌 The loop containing \(R_1\) and \(R_2\): \(18-8-V_2=0\Rightarrow V_2=10\,V\).

📌 The branch of \(R_3\) and \(R_4\) is parallel to \(R_2\), so their voltages add up to \(10\,V\): \(V_3+V_4=10\Rightarrow V_4=10-6=4\,V\).

📌 A short circuit occurs when there is a path of zero resistance between the source terminals, so the current becomes very large.

📌 In figure B the wire bridges the bottom branch (a single resistor), leaving a zero-resistance branch in parallel with the others: the equivalent resistance is zero, a short circuit. In A, C and D the wires bypass only one resistor of a two-resistor branch, so every branch keeps some resistance.

📌 To convert a galvanometer into an ammeter we connect a very small resistance (a shunt) in parallel with it so most of the current passes through it.

📌 Figure D has a \(0.1\,\Omega\) resistor in parallel with the galvanometer. (A large series resistance, such as \(200\,k\Omega\), converts it into a voltmeter.)

📌 The conventional current leaves the positive terminal (the longer line in the battery symbol) and returns to the negative one.

📌 In circuit D the positive terminal is on the right so the current passes through the ammeter to \(E\), then through the resistor to \(K\), then the switch; the voltmeter is in parallel with the resistor and the ammeter in series (correct connection).

📌 The ammeter in series with the lamp (in its branch) and the voltmeter in parallel with the lamp.

📌 Only figure B does this; in A the ammeter is in the main line so it measures the total current, and in D the voltmeter is not across the lamp.

📌 Burning out \(K\) opens the circuit so \(I=0\).

📌 No current flows in \(M\) so there is no voltage drop across it: \(V_M=0\). The voltmeter of \(K\) is now connected across the battery through \(M\) (no drop) so it reads the source voltage \(V_K=8\,V\).

📌 The energy is \(E=P\,t\) with the same time and the same voltage, so the power \(P=\dfrac{V^{2}}{R}\) is inversely proportional to the resistance.

📌 Lamp A has the smallest resistance \(5\,\Omega\) so the largest power \(\dfrac{36}{5}=7.2\,W\) and the greatest energy.

📌 The batteries oppose each other so the net emf is \(12-4=8\,V\).

📌 The loop rule: \(8-V_1-V_2=0\Rightarrow V_2=8-6=2\,V\).

📌 Superconductors lose their electrical resistance at very low temperatures; the first one discovered was mercury at \(4.2\,K\).

📌 Ohm’s law: \(I=\dfrac{V}{R}=\dfrac{12}{4}=3\,A\).

📌 The voltmeter measures the potential difference and is connected in parallel with the element (very large resistance); the ammeter measures the current in series.

\(R_{eq}=2+3+5=10\,\Omega\).

\(R_{eq}=\dfrac{6\times3}{6+3}=\dfrac{18}{9}=2\,\Omega\) (smaller than the smallest resistor).

\(I=\dfrac{12}{4+8}=1\,A\).

\(V=IR=1\times8=8\,V\).

📌 Each branch has the same voltage \(12\,V\): \(I_1=\dfrac{12}{6}=2\,A\) and \(I_2=\dfrac{12}{12}=1\,A\).

📌 The total current is \(2+1=3\,A\).

📌 \(P=\dfrac{V^{2}}{R}\Rightarrow R=\dfrac{V^{2}}{P}=\dfrac{120^{2}}{60}=240\,\Omega\).

\(R_{23}=\dfrac{6\times3}{9}=2\,\Omega\) then \(R_{eq}=4+2=6\,\Omega\).

\(I=\dfrac{12}{6}=2\,A\) so \(V_{23}=2\times2=4\,V\), which is the voltage across \(R_3\).

📌 Adding a new parallel branch decreases the equivalent resistance; the voltage is constant so the current \(I=V/R_{eq}\) increases (by the current of the new branch).

📌 The resistances: \(R_{60}=\dfrac{120^{2}}{60}=240\,\Omega\) and \(R_{100}=\dfrac{120^{2}}{100}=144\,\Omega\).

📌 In series the current is the same and the power \(P=I^{2}R\) is larger for the larger resistance, so the \(60\,W\) lamp is brighter (the opposite of the parallel case).

📌 Voltage divider: \(V=\dfrac{300}{100+300}\times20=15\,V\).

📌 Each branch: \(6+6=12\,\Omega\).

📌 The two identical branches in parallel: \(R_{eq}=\dfrac{12}{2}=6\,\Omega\).

📌 The current divides inversely with the resistance: \(I_3=8\times\dfrac{6}{3+6}=5.33\,A\).

📌 Check: \(V=\dfrac{3\times6}{9}\times8=16\,V\) so \(I_3=\dfrac{16}{3}=5.33\,A\) and \(I_6=2.67\,A\), summing to \(8\,A\).

\(R_{23}=\dfrac{20\times30}{50}=12\,\Omega\) and \(R_{eq}=8+12=20\,\Omega\).

\(I=\dfrac{20}{20}=1\,A\) then \(V_{23}=1\times12=12\,V\) so \(I_3=\dfrac{12}{30}=0.4\,A\).

📌 For identical resistors in parallel \(R_{eq}=\dfrac{R}{n}\).

\(2=\dfrac{10}{n}\Rightarrow n=5\).

📌 (a) The unit of resistance is the ohm \(\Omega\).

📌 (b) The resistor controls the current according to Ohm’s law \(V=IR\): the larger its value the smaller the current at the same voltage.

📌 (c) The four points lie on a straight line through the origin (the drawing below).

📌 (d) The slope \(\dfrac{\Delta V}{\Delta I}=\dfrac{8-0}{0.4-0}=20\,\Omega\) is the resistance.

📌 (e) \(I=\dfrac{6}{20}=0.3\,A\) so \(P=VI=6\times0.3=1.8\,W\).

📌 (f) No; the lamp curve is not a straight line because the filament resistance rises as its temperature rises, so the current is not proportional to the voltage (a non-ohmic device).

📌 (a) The resistor. It depends on the length of the conductor, its cross-section, the material and the temperature: \(R=\rho\dfrac{L}{A}\).

📌 (b) When sweating the skin becomes moister and its resistance falls, so the current through the body increases (which is more dangerous).

📌 (c) \(R=\dfrac{V}{I}=\dfrac{20}{5}=4\,\Omega\) and \(P=VI=20\times5=100\,W\).

📌 (d) \(R_2=2\,\Omega\) so \(I_2=\dfrac{20}{2}=10\,A\) and \(P_2=20\times10=200\,W\): the power doubles because the current doubles when the resistance is halved at the same voltage.

📌 (a) The current has only one path, so the same current passes through all the resistors.

📌 (b) \(R_{eq}=3+9+18=30\,\Omega\), larger than any single resistor.

📌 (c) \(I=\dfrac{30}{30}=1\,A\) in each resistor (the same).

📌 (d) \(V_1=3\,V,\ V_2=9\,V,\ V_3=18\,V\), summing to \(30\,V\): the voltage is shared in proportion to the resistances.

📌 (a) \(\dfrac{1}{R_{eq}}=\dfrac18+\dfrac1{10}+\dfrac1{15}=\dfrac{15+12+8}{120}=\dfrac{35}{120}\) so \(R_{eq}=\dfrac{120}{35}\approx3.43\,\Omega\).

📌 (b) The voltage \(18\,V\) is across every branch: \(I_1=\dfrac{18}{8}=2.25\,A\), \(I_2=\dfrac{18}{10}=1.8\,A\), \(I_3=\dfrac{18}{15}=1.2\,A\).

📌 The main ammeter \(I=2.25+1.8+1.2=5.25\,A\) (check: \(18/3.43=5.25\)).

📌 (a) \(R_{12}=\dfrac{6\times12}{6+12}=4\,\Omega\) and \(R_{eq}=4+2=6\,\Omega\).

📌 (b) \(V=IR_{eq}=3\times6=18\,V\).

📌 (c) \(V_{12}=3\times4=12\,V\) so \(I_1=\dfrac{12}{6}=2\,A\).

📌 Originally: the branch \(R_3+R_4=60\,\Omega\) in parallel with \(R_2=60\,\Omega\) gives \(30\,\Omega\), then \(R_{eq}=30+30=60\,\Omega\) and \(I=\dfrac{120}{60}=2\,A\) ✔.

📌 (a) A break in \(R_4\) opens the branch of \(R_3,R_4\) so only the branch of \(R_2\) remains: \(R_{eq}=30+60=90\,\Omega\) and \(I=\dfrac{120}{90}=1.33\,A\).

📌 (b) Shorting \(R_4\) leaves that branch with \(40\,\Omega\) only: \(\dfrac{40\times60}{100}=24\,\Omega\) then \(R_{eq}=30+24=54\,\Omega\) and \(I=\dfrac{120}{54}=2.22\,A\).

📌 (c) Shorting \(R_2\) makes the parallel group zero: \(R_{eq}=30\,\Omega\) and \(I=\dfrac{120}{30}=4\,A\).

📌 (a) The voltage across the parallel pair \(V_{23}=I_3R_3=2\times6=12\,V\) so \(I_2=\dfrac{12}{8}=1.5\,A\).

📌 The ammeter is in series with the group, so it reads the total current: \(I=I_2+I_3=1.5+2=3.5\,A\).

📌 (b) The voltmeter is across \(R_1\): \(V_1=IR_1=3.5\times2=7\,V\).

📌 The voltage of the branches \(V=I_2R_2=3\times6=18\,V\).

📌 (a) \(I_1=\dfrac{18}{9}=2\,A\).

📌 (b) \(I_3=6-3-2=1\,A\).

📌 (c) \(R_3=\dfrac{18}{1}=18\,\Omega\).

📌 (a) \(R_{12}=4\,\Omega\) and \(R_{eq}=4+2=6\,\Omega\).

📌 \(I=\dfrac{12}{6}=2\,A\).

📌 (b) \(P_3=I^{2}R_3=2^{2}\times2=8\,W\).

📌 (c) \(V_{12}=2\times4=8\,V\) so \(I_1=\dfrac{8}{6}=1.33\,A\).

📌 (a) The first branch \(R_1+R_2=20\,\Omega\) in parallel with \(R_3=20\,\Omega\): \(R_{eq}=10\,\Omega\).

\(I=\dfrac{12}{10}=1.2\,A\gt 1\,A\): the fuse would blow, so it is not suitable.

📌 (b) \(R_4\) must be in series with the fuse and the whole circuit (between \(C\) and \(D\)): the total resistance rises to \(15\,\Omega\) and the current becomes \(\dfrac{12}{15}=0.8\,A\lt 1\,A\). Between \(A\) and \(B\) it would be in the branch of \(R_3\) only: \(\dfrac{20\times25}{45}=11.1\,\Omega\) and the current \(1.08\,A\gt 1\,A\), which is not enough.

📌 (c) No; a fuse in one branch carries only that branch current (\(\dfrac{12}{20}=0.6\,A\)) and does not blow even if the total current exceeds \(1\,A\); the fuse must be in the main line to carry the total current.