28 Jul 2025
AC Circuits, Bridges, and the Transformer
Complex impedance, series and parallel LCR resonance, power, quality factor, bandwidth, AC and resistance bridges, transformer models, vector diagram, and losses.
Represent a sinusoid by the real part of a complex phasor:
\[v(t)=\Re\!\left(\sqrt2\,V e^{j\omega t}\right), \qquad i(t)=\Re\!\left(\sqrt2\,I e^{j\omega t}\right),\]where $V$ and $I$ are rms phasors and $j^2=-1$. Differentiation becomes multiplication by $j\omega$. The element impedances are
\[\boxed{Z_R=R}, \qquad \boxed{Z_L=j\omega L}, \qquad \boxed{Z_C=\frac{1}{j\omega C}=-\frac{j}{\omega C}}.\]Their reactances are $X_L=\omega L$ and $X_C=-1/(\omega C)$. Kirchhoff’s laws retain their algebraic form for phasors:
\[\boxed{\sum_{\text{node}} I=0}, \qquad \boxed{\sum_{\text{loop}} V=0}.\]They apply when all sources have the same angular frequency and the circuit has reached sinusoidal steady state.
Series LCR circuit
Kirchhoff’s loop law gives
\[V=IR+Ij\omega L+\frac{I}{j\omega C}.\]Therefore
\[\boxed{Z_s=R+j\left(\omega L-\frac{1}{\omega C}\right)}\]and
\[\lvert Z_s\rvert=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}.\]The rms current and phase angle are
\[\boxed{ I=\frac{V} {\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}}},\] \[\boxed{\tan\phi =\frac{\omega L-\dfrac{1}{\omega C}}{R}},\]where the applied voltage leads the current by $\phi$. The circuit is capacitive below resonance and inductive above resonance.
Resonance
At resonance the net reactance vanishes:
\[\omega_0L-\frac{1}{\omega_0C}=0.\]Thus
\[\boxed{\omega_0=\frac{1}{\sqrt{LC}}}, \qquad \boxed{f_0=\frac{1}{2\pi\sqrt{LC}}}.\]At $\omega_0$, $Z_s=R$, the current is maximum, and $V_L$ and $V_C$ are equal in magnitude and opposite in phase.
Average power
Let
\[v=\sqrt2V\cos\omega t, \qquad i=\sqrt2I\cos(\omega t-\phi).\]Using $2\cos A\cos B=\cos(A-B)+\cos(A+B)$,
\[p(t)=VI\left[\cos\phi+\cos(2\omega t-\phi)\right].\]The second term averages to zero over a cycle, so
\[\boxed{P_{\mathrm{av}}=VI\cos\phi}.\]Only the resistor dissipates average power:
\[\boxed{P_{\mathrm{av}}=I^2R=\frac{V^2R}{\lvert Z_s\rvert^2}}.\]The factor $\cos\phi=R/\lvert Z_s\rvert$ is the power factor.
Half-power frequencies, bandwidth, and quality factor
At resonance $I_0=V/R$. At either half-power frequency, the current is $I_0/\sqrt2$, so
\[R^2+\left(\omega L-\frac{1}{\omega C}\right)^2=2R^2.\]Hence
\[\left\lvert\omega L-\frac{1}{\omega C}\right\rvert=R.\]For the lower frequency,
\[L\omega_1^2+R\omega_1-\frac1C=0,\]which gives the positive root
\[\omega_1=\frac{-R+\sqrt{R^2+4L/C}}{2L}.\]For the upper frequency,
\[L\omega_2^2-R\omega_2-\frac1C=0,\]so
\[\omega_2=\frac{R+\sqrt{R^2+4L/C}}{2L}.\]Subtracting and multiplying the roots gives
\[\boxed{\Delta\omega=\omega_2-\omega_1=\frac{R}{L}}, \qquad \boxed{\omega_1\omega_2=\omega_0^2}.\]The series quality factor is
\[\boxed{ Q_s=\frac{\omega_0}{\Delta\omega} =\frac{\omega_0L}{R} =\frac{1}{\omega_0CR}}.\]At resonance, $Q_s=\lvert V_L\rvert/V=\lvert V_C\rvert/V$.
Parallel LCR circuit
For ideal $R$, $L$, and $C$ branches in parallel,
\[I=V\left(\frac1R+\frac{1}{j\omega L}+j\omega C\right).\]Thus the admittance is
\[\boxed{ Y_p=\frac1R+j\left(\omega C-\frac{1}{\omega L}\right)}.\]At
\[\boxed{\omega_0=\frac1{\sqrt{LC}}},\]the inductive and capacitive branch currents cancel. The source current is minimum and
\[\boxed{Z_p=\frac1{Y_p}=R}\]is maximum.
For a fixed source current, the half-power points satisfy $\lvert Z_p\rvert=R/\sqrt2$, equivalently
\[\left\lvert\omega C-\frac{1}{\omega L}\right\rvert=\frac1R.\]The two positive roots give
\[\boxed{\Delta\omega_p=\frac{1}{RC}},\]and
\[\boxed{ Q_p=\frac{\omega_0}{\Delta\omega_p} =\omega_0RC =\frac{R}{\omega_0L}}.\]This bandwidth definition uses a current-driven parallel network; with an ideal fixed-voltage source, the resistor power $V^2/R$ itself does not vary with frequency.
General bridge balance
Let a four-arm bridge have impedances $Z_1,Z_2$ in the left branch and $Z_3,Z_4$ in the right branch. The source is across the top and bottom nodes, and a null detector joins the two midpoints. With no detector current,
\[V_B=V\frac{Z_2}{Z_1+Z_2}, \qquad V_D=V\frac{Z_4}{Z_3+Z_4}.\]Balance requires $V_B=V_D$:
\[\frac{Z_2}{Z_1+Z_2} =\frac{Z_4}{Z_3+Z_4}.\]Cross-multiplication and cancellation of $Z_2Z_4$ give
\[\boxed{Z_1Z_4=Z_2Z_3}.\]Because this is a complex equation, both magnitude and phase must balance in an AC bridge.
De-Sauty bridge
The De-Sauty bridge compares two nearly loss-free capacitances. Choose
\[Z_1=\frac{1}{j\omega C_x},\quad Z_2=R_2,\quad Z_3=\frac{1}{j\omega C_s},\quad Z_4=R_4.\]The balance condition becomes
\[\frac{R_4}{j\omega C_x} =\frac{R_2}{j\omega C_s}.\]Cancel $j\omega$ and cross-multiply:
\[R_4C_s=R_2C_x.\]Therefore
\[\boxed{\frac{C_x}{C_s}=\frac{R_4}{R_2}}.\]The result is frequency independent for ideal capacitors. Appreciable dielectric loss adds a resistive component and prevents a sharp balance with this simple bridge.
Carey Foster bridge
The Carey Foster bridge compares nearly equal resistances $X$ and $Y$. Let $r$ be the slide-wire resistance per unit length, $L_w$ its total length, $l$ a balance length measured from the left end, and $R_L,R_R$ the fixed end resistances. Choose equal ratio arms, so the two lower-arm resistances are equal at balance.
Before interchanging $X$ and $Y$,
\[X+R_L+rl_1 =Y+R_R+r(L_w-l_1).\]After interchanging them,
\[Y+R_L+rl_2 =X+R_R+r(L_w-l_2).\]Rearrange the first equation:
\[X-Y=R_R-R_L+rL_w-2rl_1.\]Rearrange the second:
\[X-Y=R_L-R_R-rL_w+2rl_2.\]Add these two expressions:
\[2(X-Y)=2r(l_2-l_1).\]Hence
\[\boxed{X-Y=r(l_2-l_1)}.\]The interchange cancels the unknown end corrections. The sign follows the stated left-end convention; the magnitude gives the resistance difference.
Anderson bridge
The Anderson bridge measures the series resistance $R_x$ and self-inductance $L_x$ of a real coil using a standard capacitor $C$.
Let
\[Z_x=R_x+j\omega L_x, \qquad Z_C=\frac{1}{j\omega C}.\]The left branch is $Z_x$ in series with $R_2$. On the right, $R_3$ is in series with the parallel combination of $R_4$ and $r+Z_C$. The detector compares the junction of $Z_x,R_2$ with the junction of $r,Z_C$.
At null, no detector current flows. The left detector-node voltage is
\[\frac{V_B}{V}=\frac{R_2}{Z_x+R_2}.\]For the right network, first form
\[Z_p=R_4\parallel(r+Z_C) =\frac{R_4(r+Z_C)}{R_4+r+Z_C}.\]If $F$ is the junction after $R_3$, then
\[\frac{V_F}{V}=\frac{Z_p}{R_3+Z_p}, \qquad \frac{V_E}{V_F}=\frac{Z_C}{r+Z_C}.\]Multiplying and simplifying,
\[\frac{V_E}{V} =\frac{R_4Z_C} {R_3R_4+r(R_3+R_4)+Z_C(R_3+R_4)}.\]Divide numerator and denominator by $Z_C$, using $1/Z_C=j\omega C$:
\[\frac{V_E}{V} =\frac{R_4} {R_3+R_4+j\omega C\left[R_3R_4+r(R_3+R_4)\right]}.\]At balance, $V_B=V_E$:
\[\frac{R_2}{R_x+R_2+j\omega L_x} = \frac{R_4} {R_3+R_4+j\omega C\left[R_3R_4+r(R_3+R_4)\right]}.\]Equate real parts after cross-multiplication:
\[R_2(R_3+R_4)=R_4(R_x+R_2),\]so
\[\boxed{R_x=\frac{R_2R_3}{R_4}}.\]Equating imaginary parts gives
\[\omega R_2C\left[R_3R_4+r(R_3+R_4)\right] =\omega R_4L_x.\]Cancel $\omega$:
\[\boxed{ L_x=\frac{CR_2}{R_4} \left[R_3R_4+r(R_3+R_4)\right]}.\]Both ideal balance equations are independent of frequency. The equivalent voltage-divider form above also makes clear that null requires both detector voltages to have the same magnitude and phase.
Transformer
A transformer uses mutual induction between windings linked by a common alternating core flux. Let
\[\Phi(t)=\Phi_m\sin\omega t.\]For a winding of $N$ turns,
\[e(t)=-N\frac{\mathrm d\Phi}{\mathrm dt} =-\omega N\Phi_m\cos\omega t.\]The peak emf is $\omega N\Phi_m$, so
\[E_{\mathrm{rms}} =\frac{\omega N\Phi_m}{\sqrt2} =\sqrt2\,\pi fN\Phi_m \simeq4.44fN\Phi_m.\]For an ideal transformer, the two windings share the same flux:
\[\boxed{\frac{V_s}{V_p} =\frac{E_s}{E_p} =\frac{N_s}{N_p}}.\]Conservation of power and opposing ampere-turns give
\[V_pI_p=V_sI_s, \qquad N_pI_p=N_sI_s\]in magnitude. Hence
\[\boxed{\frac{I_s}{I_p}=\frac{N_p}{N_s}}.\]If $a=N_p/N_s$, a load $Z_L$ referred to the primary is
\[\boxed{Z_L'=a^2Z_L}.\]Indeed,
\[\frac{V_p}{I_p} =\frac{aV_s}{I_s/a} =a^2\frac{V_s}{I_s}.\]Equivalent circuit of a real transformer
A practical winding has copper resistance $R_1,R_2$ and leakage reactance $X_1,X_2$. Core magnetization is represented by $jX_m$, while core loss is represented by $R_c$. Referred to the primary,
\[R_2'=a^2R_2,\qquad X_2'=a^2X_2,\qquad Z_L'=a^2Z_L.\]The approximate primary-referred series parameters are
\[\boxed{R_{\mathrm{eq}}=R_1+R_2'}, \qquad \boxed{X_{\mathrm{eq}}=X_1+X_2'}.\]The shunt branch $R_c\parallel jX_m$ accounts for core-loss current and magnetizing current.
Vector diagram
For a lagging secondary load, take $V_2$ as reference. The current $I_2$ lags by the load angle $\phi$. The resistive drop $I_2R_2$ is in phase with $I_2$, and the leakage-reactance drop $jI_2X_2$ leads $I_2$ by $90^\circ$. Therefore
\[\boxed{\mathbf E_2 =\mathbf V_2+\mathbf I_2R_2+j\mathbf I_2X_2}.\]With a consistent dot convention, $\mathbf E_1/\mathbf E_2=N_1/N_2$. Primary drops are added similarly to relate $\mathbf V_1$ and $\mathbf E_1$.
Transformer losses
The principal real-power losses are:
-
Copper loss
\[\boxed{P_{\mathrm{cu}}=I_1^2R_1+I_2^2R_2}.\]It varies approximately as load current squared.
-
Hysteresis loss. Repeated reversal of core magnetization dissipates the $B$-$H$ loop energy every cycle. A soft, narrow-loop core reduces it.
-
Eddy-current loss. Changing flux drives circulating currents in the conducting core. Thin insulated laminations increase the transverse resistance and reduce this loss.
-
Stray load and dielectric losses. Leakage flux induces additional currents in windings and structural metal; insulation also dissipates a smaller dielectric loss.
Leakage flux mainly causes reactive voltage drop and poor regulation; associated eddy currents contribute real stray loss. At fixed supply voltage and frequency, the core loss is approximately constant, whereas copper loss changes with load. The efficiency is
\[\boxed{ \eta=\frac{P_{\mathrm{out}}} {P_{\mathrm{out}}+P_{\mathrm{cu}}+P_{\mathrm{core}}+P_{\mathrm{stray}}}}.\]
The series and parallel half-power roots, quality factors, bridge balances, Anderson reduction, rms transformer emf, turns-current ratios, and referred impedances are verified with exact zero residuals in the Unit III AC, bridges, and transformer worksheet.
Discussion