A $5\,\text{V}$ battery in series with a $25\,\Omega$ resistor is connected between a left node-rail and a right node-rail, driving three parallel branches between the two rails (the battery branch carries the current $I_1$). Taking the left rail as the higher-potential side: the top branch ($I_4$) has a diode oriented to conduct from the left rail to the right rail in series with a $125\,\Omega$ resistor; the middle branch ($I_3$) has a diode oriented the OPPOSITE way (blocking current from left to right) in series with a $125\,\Omega$ resistor; the third branch ($I_2$) has a diode oriented to conduct from left to right in series with a $125\,\Omega$ resistor. Each diode has a forward-bias resistance of $25\,\Omega$ and infinite resistance in reverse bias. Find the values of the currents $I_1$, $I_2$, $I_3$ and $I_4$.

A 5V battery in series with a 25Ω resistor is connected between a left node-rail and a right node-rail, drivin

NCERT Exemplar · Q17

5\,\text{V}

Karnataka PUCLong· 5mImportance★★★★★

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Concept understanding — Diode Resistance Calculation

Diode Resistance: What Does It Even Mean?

Think of a diode as a one-way valve for electricity. When you push current through it in the forward direction, the diode doesn't just let everything through freely — it resists the flow, just like any other component. But here's the twist: that resistance isn't a fixed number like a resistor's 100 Ω. It changes depending on how much voltage you apply.

Why? Because a diode is a semiconductor device. Its current-voltage relationship follows the Shockley equation:

I = I_S (e^{V / \eta V_T} - 1)

where $I_S$ is the reverse saturation current, $V$ is the applied voltage, $\eta$ is the ideality factor (usually 1 for silicon), and $V_T \approx 26 \text{ mV}$ at room temperature.

This exponential curve means that a tiny change in voltage can cause a huge change in current. So the "resistance" you measure depends entirely on where you are on that curve.

Two Kinds of Diode Resistance

Because the I-V curve is nonlinear, we define two different resistances — each useful in different situations.

1. Static (DC) Resistance

This is the simplest idea: just apply Ohm's law using the total voltage and total current at a given operating point.

R_{DC} = \frac{V}{I}

For example, if a diode has 0.7 V across it and 10 mA flowing through it, its DC resistance is:

R_{DC} = \frac{0.7}{0.01} = 70 \ \Omega

Note

Static resistance tells you the average opposition to current at that specific point. It's useful for power calculations ( $P = I^2 R_{DC}$ ) but not for small signal analysis.

2. Dynamic (AC) Resistance

This is the more important one for circuit design. It tells you how the diode responds to small changes in voltage around a fixed operating point.

Mathematically, dynamic resistance is the slope of the I-V curve at that point:

r_d = \frac{dV}{dI}

For a forward-biased diode obeying the Shockley equation, we can derive a clean formula. Starting from:

I = I_S e^{V / \eta V_T}

(ignoring the -1, which is negligible in forward bias)

Differentiate:

\frac{dI}{dV} = \frac{I}{\eta V_T}

Therefore:

r_d = \frac{dV}{dI} = \frac{\eta V_T}{I}

r_d = \frac{\eta V_T}{I}

At room temperature with $\eta = 1$ and $V_T = 26 \text{ mV}$ :

r_d = \frac{26 \text{ mV}}{I}

So if the diode current is 10 mA, $r_d = 2.6 \ \Omega$ — much smaller than the 70 Ω DC resistance.

Watch out

Dynamic resistance is not a physical resistor inside the diode. It's a small-signal model parameter. You cannot use it with DC voltages or large signals — only for tiny variations around the operating point.

When Do You Use Each?

Situation	Use
Finding DC power dissipation	$R_{DC}$
Designing a biasing circuit	$R_{DC}$
Analyzing small-signal amplifier response	$r_d$
Calculating voltage regulation in a Zener diode	$r_d$ (called Zener impedance)

A Quick Example to Tie It Together

A silicon diode ( $\eta = 1$ ) is forward biased with $V = 0.7 \text{ V}$ and carries $I = 20 \text{ mA}$ .

Static resistance: …

Why this formula?

Diode Resistance: Why It Changes with Operating Point

A diode is not a linear resistor. Its current-voltage relationship follows the Shockley equation:

I = I_S (e^{V / \eta V_T} - 1)

where $I_S$ is the reverse saturation current, $\eta$ is the ideality factor (typically 1–2), and $V_T = kT/q \approx 26\,\text{mV}$ at room temperature.

Because the $I$ – $V$ curve is exponential, the diode's resistance depends entirely on where you are on that curve. There are two distinct resistances we care about: DC resistance (static) and AC resistance (dynamic).

DC Resistance (Static Resistance)

Definition: The ratio of the DC voltage across the diode to the DC current through it at a given operating point.

R_{DC} = \frac{V}{I}

Why this formula? It's simply Ohm's law applied to the DC values. If you put 0.7 V across a diode and get 10 mA through it, the DC resistance is $0.7 / 0.01 = 70\,\Omega$ . But this number is misleading — it doesn't tell you how the diode responds to a small change in voltage.

Watch out

DC resistance is rarely useful in circuit analysis because diodes are never operated as fixed resistors. It's just a snapshot at one point.

AC Resistance (Dynamic Resistance)

Definition: The slope of the $I$ – $V$ curve at a given operating point — i.e., the ratio of a small change in voltage to the resulting small change in current.

r_d = \frac{dV}{dI}

Why this formula? For small signals (like an AC voltage superimposed on a DC bias), the diode behaves approximately linearly around that bias point. The dynamic resistance is the local slope of the $I$ – $V$ curve.

Now let's derive the actual expression.

Derivation of $r_d = \frac{\eta V_T}{I}$

Start from the Shockley equation. For forward bias where $V \gg V_T$ , the $-1$ term is negligible:

I \approx I_S e^{V / \eta V_T}

Take the derivative with respect to $V$ :

\frac{dI}{dV} = I_S \cdot \frac{1}{\eta V_T} \cdot e^{V / \eta V_T} = \frac{I}{\eta V_T}

The dynamic resistance is the reciprocal:

r_d = \frac{dV}{dI} = \frac{\eta V_T}{I}

r_d = \frac{\eta V_T}{I}

Key insight: The dynamic resistance is inversely proportional to the DC current $I$ . At higher currents, the diode's $I$ – $V$ curve is steeper, so a small voltage change produces a larger current change — meaning lower resistance.

Why This Matters …

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The middle diode is reverse biased, so its branch carries no current: $I_3=0$ . Each of the other two branches presents $150\,\Omega$ ( $125\,\Omega$ resistor + $25\,\Omega$ forward diode); with the $25\,\Omega$ battery resistor, the $5\,\text{V}$ source drives $I_1=0.05\,\text{A}$ , which splits equally to give $I_2=I_4=0.025\,\text{A}$ .

Which branches conduct

The $5\,\text{V}$ battery (with its series $25\,\Omega$ ) drives current out of the left rail. The diodes decide the paths:

Top branch ( $I_4$ ) and third branch ( $I_2$ ): diodes forward biased $\Rightarrow$ conduct.
Middle branch ( $I_3$ ): diode reverse biased $\Rightarrow$ open, so $I_3=0$ .

Branch resistance

Each conducting branch = resistor + forward diode:

R_{branch}=125+25=150\,\Omega.

The two conducting branches are in parallel:

R_p=\frac{150\times150}{150+150}=75\,\Omega.

Total current from the battery …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set D-41 markMCQ
$D_1$
›Reveal solutionSolution
The two diodes face opposite ways, so only one $30\,\Omega$ branch conducts; the circuit is then simply $30\,\Omega + 70\,\Omega$ in series with the 3 V cell.

Step 1 — What an ideal diode does.

An ideal diode is a perfect one-way switch:

Forward biased (conventional current entering the anode, i.e. flowing in the direction the triangle points): resistance $= 0$ — behaves as a plain wire.

Reverse biased: resistance $= \infty$ — behaves as an open circuit (a break in the wire).

Step 2 — Read the circuit.

There are three branches between the left rail and the right rail:

Top: $D_1$ in series with $30\,\Omega$ . $D_1$ points left, so it can pass conventional current only right → left.

Middle: $D_2$ in series with $30\,\Omega$ . $D_2$ points right, so it can pass current only left → right.

Bottom: the 3 V cell in series with the $70\,\Omega$ resistor.

The two diode branches are in parallel with each other, and that parallel combination is in series with the cell and the $70\,\Omega$ resistor around the loop.

Step 3 — The key observation: exactly one diode conducts.

The cell drives current one way round the loop, so the current in the parallel section must flow either left→right or right→left. Because $D_1$ and $D_2$ are wired in opposite senses:

If the current in that section is left → right, then $D_2$ is forward biased (conducts) and $D_1$ is reverse biased (blocks).

If it is right → left, then $D_1$ conducts and $D_2$ blocks.

Either way, exactly one of the two $30\,\Omega$ branches carries current and the other is an open circuit. The two $30\,\Omega$ resistors are therefore NOT in parallel — that is the trap the question is built around. (If you wrongly parallel them you get $15 + 70 = 85\,\Omega$ and $I \approx 0.035\,$ A, which matches no option.)

Step 4 — Reduce the circuit.

The conducting diode is an ideal short (0 Ω), so the whole loop is just: …
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KCET 2023Set A-31 markMCQ
$4.7~\mathrm{k}\Omega \pm 5\%$
›Reveal solutionSolution
Write $4.7\ \mathrm{k\Omega}$ as $47 \times 10^2\ \Omega$ ; the third band is the multiplier $10^2$ , which is red.

Step 1 — The colour code

For a four-band carbon resistor:

Band 1 → first significant digit

Band 2 → second significant digit

Band 3 → multiplier (power of ten)

Band 4 → tolerance (gold $=\pm5\%$ , silver $=\pm10\%$ )

Digit/multiplier colours: Black $0$ , Brown $1$ , Red $2$ , Orange $3$ , Yellow $4$ , Green $5$ , Blue $6$ , Violet $7$ , Grey $8$ , White $9$ .

Step 2 — Express the value in "two digits × multiplier" form
$R = 4.7\ \mathrm{k\Omega} = 4700\ \Omega = 47 \times 10^{2}\ \Omega$
Step 3 — Read off the bands

Band 1 = digit $4$ → yellow

Band 2 = digit $7$ → violet

Band 3 = multiplier $10^{2}$ → the colour for $2$ → red …
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KCET 2023Set A-31 markMCQ
$82~\Omega \pm 10\%$
›Reveal solutionSolution
Read the four bands as (digit)(digit)(multiplier)(tolerance): gray $8$ , red $2$ , gold $\times10^{-1}$ , gold $\pm5\%$ .

Step 1 — The colour code.

In a four-band resistor:

Band 1 = first significant digit

Band 2 = second significant digit

Band 3 = decimal multiplier

Band 4 = tolerance

The digit values (B B ROY of Great Britain has a Very Good Wife): Black 0, Brown 1, Red 2, Orange 3, Yellow 4, Green 5, Blue 6, Violet 7, Gray 8, White 9.

Step 2 — Read the bands.

Band Colour Meaning
1 Gray $8$
2 Red $2$
3 Gold multiplier $\times 10^{-1}$
4 Gold tolerance $\pm 5\%$
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COMEDK 2023Set 2023-E1 markMCQ
$20 \mu \mathrm{A}$
›Reveal solutionSolution

The reverse resistance is $\Delta V/\Delta I = 5\,\text{V}/20\,\mu\text{A} = 250\,\text{k}\Omega$ .

The dynamic reverse resistance of the diode is the ratio of the change in reverse potential to the resulting change in reverse (leakage) current.
$\Delta V = 15 - 10 = 5\ \text{V}$
$\Delta I = 40 - 20 = 20\ \mu\text{A} = 20\times10^{-6}\ \text{A}$ …
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KCET 2018Set A-11 markMCQ
$(X_L)$
›Reveal solutionSolution

$X_L = 2\pi\nu L$ is a linear law through the origin, so the $X_L$ – $\nu$ graph is a straight line from the origin.

Step 1 — Where inductive reactance comes from.

For a pure inductor driven by $i = i_0\sin\omega t$ , the back-emf is $L\,di/dt = \omega L\, i_{0}\cos\omega t$ . Comparing the amplitude of the voltage with that of the current, the effective 'resistance' offered by the inductor is
$X_{L} = \omega L = 2\pi \nu L .$
Physically: a faster-changing current induces a larger opposing emf, so the inductor obstructs high frequencies more.

Step 2 — Identify the functional form.

With $L$ a constant of the coil,
$X_{L} = (2\pi L)\,\nu ,$
which is of the form $y = mx$ with $m = 2\pi L$ and zero intercept. So $X_L \propto \nu$ : a straight line passing through the origin.

Step 3 — Check the limits. …
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Band	Colour	Meaning
1	Gray	$8$
2	Red	$2$
3	Gold	multiplier $\times 10^{-1}$
4	Gold	tolerance $\pm 5\%$

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Which branches conduct

Branch resistance

Total current from the battery …

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