Exercises — Read and interpret circuit schematic symbols

Reading left to right on the figure:

• (a) Battery / cell — the long thin line is $+$, the short thick line is $-$. Job: maintains a fixed EMF $\mathcal{E}$ (defined just above — an electrical "pump" that pushes current out of $+$).
• (b) Resistor (zigzag) — obeys Ohm's Law, $V=IR$. Job: drops voltage / limits current.
• (c) Capacitor (two equal parallel plates) — stores charge, $Q=CV$. Job: blocks steady DC after charging (you will apply this in L2.4 and L5.2).
• (d) LED / diode (triangle + bar) — one-way current valve; the triangle points with conventional current. Job: conducts one way (LED also emits light).
• (e) Ground — the $0\text{ V}$ reference node. Job: the point every voltage is measured against.
• (f) Switch — a gap that can open or close. Open = the node is broken (no current, open circuit); closed = an ideal wire ($0\text{ V}$ drop). Job: turn a branch on or off.

• (i) Dot present (+ with dot) → the wires are connected — they form one node (one shared potential).
• (ii) No dot (+ crossover) → the wires cross over — they are two separate nodes; the lines just pass, like one road bridging over another.
• (iii) T-junction → a wire ending into the middle of another is a connection, even without a drawn dot. A tee can only mean "join" (the ending wire has nowhere else to go), so the T-shape implies a node by convention. Many schematics still add a dot for clarity, but the tee alone already means connected.

Rule of thumb: dot or tee = joined; bare + crossover = not joined.

The resistor symbol means Ohm's Law $V=IR$. Why this law and not another? The whole point of a resistor is a straight-line relationship between the voltage across it and the current through it — that is Ohm's law. $V=IR=0.03\times 220=6.6\text{ V}$ So $\boxed{6.6\text{ V}}$ is dropped across it.

The shape tells you which quantity is shared (see Series and Parallel Circuits).

(a) Series — same current through both, so resistances add: $R_{eq}=R_1+R_2=300+600=900\Omega$

(b) Parallel — same voltage across both, so reciprocals add: $\frac{1}{R_{eq}}=\frac{1}{300}+\frac{1}{600}=\frac{2}{600}+\frac{1}{600}=\frac{3}{600}$ $R_{eq}=\frac{600}{3}=200\Omega$ Notice $200\Omega$ is less than either resistor: parallel gives current two paths, so charge flows more easily → lower resistance.

A diode conducts only when its triangle points with conventional current (from $+$ toward $-$). On the top wire, clockwise current flows left → right, so the triangle must point right (bar on the right). Point it the other way and the LED is dark — see Diodes and LEDs.

The capacitor symbol means the law $Q=CV$. Why this law and not Ohm's? A capacitor doesn't resist current — it accumulates charge on its two plates until the voltage across it matches what's pushing it. So the relevant quantity is stored charge, not a $V$–$I$ ratio. $Q=CV=(100\times 10^{-6})\times 9=9\times 10^{-4}\text{ C}=0.9\text{ mC}$ Steady current: once fully charged, the plates are at the full $9\text{ V}$ and no more charge can flow onto them, so a capacitor blocks steady DC: $I_{\text{steady}}=0\text{ A}$ So $Q=0.9\text{ mC}$ and, in the steady state, no current passes through it.

Step 1 — apply Kirchhoff's Voltage Law. Around one loop, the voltages sum to zero: the battery hands out $\mathcal{E}$, the LED takes $V_{LED}$, the resistor takes the rest. $\mathcal{E}-V_{LED}-IR=0\Rightarrow I=\frac{\mathcal{E}-V_{LED}}{R}$ Why this formula? In a series loop the current is the same everywhere (charge can't pile up), so the resistor must drop exactly whatever the battery didn't hand to the LED. $I=\frac{9-2}{330}=\frac{7}{330}\approx 0.0212\text{ A}=21.2\text{ mA}$

Step 2 — power in the resistor. Power $=I^{2}R$. Why this formula? As current $I$ pushes through the resistance $R$, charges collide inside the material and dump energy as heat (Joule heating). Each collision loss scales with how hard you push ($V_R=IR$) times how much charge flows per second ($I$), so $P=V_R\cdot I=(IR)\cdot I=I^{2}R$. That is why we use $I^{2}R$ here and not some other product. $P=I^{2}R=(0.0212{)}^2\times 330\approx 0.148\text{ W}$ So $I\approx 21.2\text{ mA}$ and $P\approx 0.15\text{ W}$.

Walk the loop and cut at each component:

1. Node A — from battery $+$ up to the resistor's top lead.
2. Node B — between the resistor and the LED.
3. Node C — from the LED's far side back to battery $-$.

That's 3 nodes. Every point on node A sits at the same potential (see Electric Potential and Voltage); the drops happen only across the components between nodes.

A dot means the $+$ wire and the $-$ wire are the same node. That joins the battery terminals directly with (ideally) zero resistance → a short circuit. Current is limited only by tiny wire/internal resistance, so it spikes dangerously. The dot here is a bug, not a feature. (In the figure, follow the red loop: it goes straight from $+$ back to $-$ through the dot, bypassing everything.)

Design backwards from KVL. The resistor must absorb the leftover voltage at the target current: $V_R=\mathcal{E}-V_{LED}=5-2=3\text{ V},R=\frac{V_R}{I}=\frac{3}{0.02}=150\Omega$ So a $\boxed{150\Omega }$ resistor sets the LED to $20\text{ mA}$. (This is the parent note's example, solved forwards here as a design.)

Two $200\Omega$ in series give $400\Omega$ (too big). Two in parallel give $\frac{1}{R_{eq}}=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}\Rightarrow R_{eq}=100\Omega .$ Neither hits $120$ exactly with two resistors. Parallel ($100\Omega$) is closest — it's the right layout because we need a value below $200\Omega$, and only parallel drops below the smallest resistor. (To land exactly on $120$ you'd add a third resistor; two can't do it.)

Attach the ground symbol to that node. Ground doesn't force the physics to zero — it declares that node the reference, so all other voltages are measured relative to it (see Grounding and Reference Voltages). It's a bookkeeping choice: you name one node "$0\text{ V}$" and every other potential becomes a difference from it.

Step 1 — collapse the parallel pair. They share both nodes, so: $R_{23}=\frac{1}{\frac{1}{200}+\frac{1}{200}}=100\Omega .$ Step 2 — now it's a simple series loop: $R_1+R_{23}=100+100=200\Omega .$ Step 3 — total current from the battery (Ohm's law on the whole loop): $I_{\text{total}}=\frac{12}{200}=0.06\text{ A}=60\text{ mA}.$ Step 4 — voltage across the parallel block. All $60\text{ mA}$ flows through $R_1$ first, dropping $V_1=I_{\text{total}}{R}_1=0.06\times 100=6\text{ V}$. The parallel block gets the remaining $12-6=6\text{ V}$. Step 5 — current through $R_2$ (its own Ohm's law, since parallel branches share voltage): $I_2=\frac{6}{200}=0.03\text{ A}=30\text{ mA}.$ Check: $R_3$ also carries $30\text{ mA}$, and $30+30=60\text{ mA}$ — matches $I_{\text{total}}$. ✓

Step 1 — what does a capacitor do to steady DC? From $Q=CV$: once charged, no more charge can move onto the plates, so a capacitor blocks steady DC — the $R_3$–$C$ branch carries zero current in the steady state. $I_{R_3}=0\text{ A}$ Step 2 — so the branch reduces the circuit. With no current in the $R_3$ branch, only $R_2$ carries current. The loop is now just $R_1+R_2=100+200=300\Omega$, giving $I=12/300=40\text{ mA}$ through $R_1$ and $R_2$. Step 3 — voltage across the capacitor. With $I_{R_3}=0$, there is no drop across $R_3$ ($V=IR=0$). So the capacitor sees the full voltage across the parallel node pair. That node pair voltage is $\mathcal{E}-I{R}_1=12-0.04\times 100=12-4=8\text{ V}$. $V_C=8\text{ V},Q=CV=(47\times 10^{-6})(8)=3.76\times 10^{-4}\text{ C}=0.376\text{ mC}$ So $I_{R_3}=0$, $V_C=8\text{ V}$, and the capacitor holds $0.376\text{ mC}$.

A diode is a one-way valve. Pointed backwards, it blocks current: the branch is effectively an open circuit. $I=0\text{ A},V_R=IR=0\text{ V}.$ With no current, the resistor drops nothing — so by KVL the entire $9\text{ V}$ must appear across the (non-conducting) LED: $V_{LED\text{ terminals}}=\mathcal{E}-V_R=9-0=9\text{ V}.$ The LED stays dark. This is the "every case" check the parent note demands: forward → lights, reverse → open, full EMF stranded across the blocking element.

An open switch breaks the node → open circuit → $I=0$. With $I=0$, the resistor drops $IR=0\text{ V}$ and the LED drops $0\text{ V}$ (no current to conduct). By KVL the whole EMF must appear somewhere — it lands across the open switch: $V_{\text{switch}}=\mathcal{E}=9\text{ V}.$ So: no current flows, yet the open switch reads the full $9\text{ V}$. Closing it makes it an ideal wire ($0\text{ V}$ drop) and current resumes.

• Series of $300$ and $600\Omega$? ::: $900\Omega$
• Same two in parallel? ::: $200\Omega$
• Resistor for a $2\text{ V}$ LED at $20\text{ mA}$ off $5\text{ V}$? ::: $150\Omega$
• Charge on a $100\mu \text{F}$ cap at $9\text{ V}$? ::: $0.9\text{ mC}$
• Voltage across an open switch in a $9\text{ V}$ loop? ::: $9\text{ V}$ (full EMF)
• Current through a reverse-biased LED? ::: $0\text{ A}$
• Does a T-junction (no dot) mean connected? ::: Yes — a tee implies a connection.