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Current Electricity - Limitations of Ohm’s Law

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Ohm's Law states that the current II flowing through a conductor is directly proportional to the potential difference VV across its ends, provided physical conditions like temperature remain constant. Devices that follow this linear relationship (V∝IV \propto I) are called ohmic conductors.

A linear V-I graph passing through the origin representing an ohmic conductor.
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Non-ohmic behavior occurs when the V−IV-I relationship is non-linear. One common limitation is the heating effect: as current increases, the temperature of a conductor rises, causing its resistance to increase. Consequently, the graph curves away from the straight line at high currents.

V-I graph showing a straight line at low current that curves upwards at higher values due to temperature increase.
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The relationship between VV and II may depend on the sign of VV. In devices like a p-n junction diode, a small forward voltage allows significant current, but reversing the polarity results in very little current until breakdown occurs.

V-I characteristic of a diode showing asymmetry between positive and negative voltage regions.
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Some materials, like Gallium Arsenide (GaAs), exhibit a region where current decreases as voltage increases. This is known as the 'negative resistance' region, which is a significant departure from Ohm's Law.

Graph for GaAs showing a peak followed by a downward slope, indicating negative resistance.

📐Formulae

V=IRV = IR

rd=ΔVΔIr_d = \frac{\Delta V}{\Delta I}

RT=R0[1+α(T−T0)]R_T = R_0[1 + \alpha(T - T_0)]

P=I2R=V2RP = I^2R = \frac{V^2}{R}

💡Examples

Problem 1:

In a certain semiconductor device, the current increases from 2 mA2\text{ mA} to 10 mA10\text{ mA} when the voltage is increased from 0.6 V0.6\text{ V} to 0.8 V0.8\text{ V}. Calculate the dynamic resistance rdr_d in this region.

Solution:

Given: ΔV=0.8 V−0.6 V=0.2 V\Delta V = 0.8\text{ V} - 0.6\text{ V} = 0.2\text{ V} and ΔI=10 mA−2 mA=8 mA=8×10−3 A\Delta I = 10\text{ mA} - 2\text{ mA} = 8\text{ mA} = 8 \times 10^{-3}\text{ A}. Using the formula for dynamic resistance: rd=ΔVΔI=0.28×10−3=2008=25 Ωr_d = \frac{\Delta V}{\Delta I} = \frac{0.2}{8 \times 10^{-3}} = \frac{200}{8} = 25\, \Omega

Explanation:

The dynamic resistance is calculated as the ratio of a small change in voltage to the resulting change in current, which is useful for non-ohmic devices where the V−IV-I slope is not constant.

Problem 2:

A student plots a V−IV-I graph for a metallic wire at two different temperatures T1T_1 and T2T_2. The graph for T2T_2 has a steeper slope (where VV is on the y-axis and II is on the x-axis) than T1T_1. Which temperature is higher?

Solution:

The slope of a V−IV-I graph represents the resistance RR. Since the slope for T2T_2 is greater than the slope for T1T_1, it implies R2>R1R_2 > R_1. For metals, resistance increases with temperature according to RT=R0(1+αΔT)R_T = R_0(1 + \alpha \Delta T). Therefore, T2>T1T_2 > T_1.

Explanation:

This demonstrates how Ohm's law fails to remain constant when temperature changes, leading to different resistance values for the same material.

Problem 3:

Identify the region in the given V−IV-I characteristic of a tunnel diode where the device exhibits negative dynamic resistance. If at point A the voltage is 0.2 V0.2 \text{ V} and current is 4 mA4 \text{ mA}, and at point B the voltage is 0.4 V0.4 \text{ V} and current is 2 mA2 \text{ mA}, calculate the dynamic resistance rdr_d for this region.

V-I graph with a peak at A and a valley at B; the slope between A and B is negative.

Solution:

  1. Negative dynamic resistance occurs where current II decreases as voltage VV increases.
  2. Change in voltage ΔV=0.4 V−0.2 V=0.2 V\Delta V = 0.4 \text{ V} - 0.2 \text{ V} = 0.2 \text{ V}.
  3. Change in current ΔI=2 mA−4 mA=−2 mA=−2×10−3 A\Delta I = 2 \text{ mA} - 4 \text{ mA} = -2 \text{ mA} = -2 \times 10^{-3} \text{ A}.
  4. Dynamic resistance rd=ΔVΔIr_d = \frac{\Delta V}{\Delta I}
  5. rd=0.2−2×10−3=−100 \Omegar_d = \frac{0.2}{-2 \times 10^{-3}} = -100 \text{ \Omega}.

Explanation:

The negative sign in the resistance value indicates that the potential difference and current changes are in opposite directions, characteristic of the negative resistance region in semiconductors.

Problem 4:

A non-ohmic resistor has a V−IV-I characteristic defined by the curve shown. If the current through the resistor is 2 A2 \text{ A}, find the dynamic resistance at that point using the graph where V=I2V = I^2.

A parabolic V-I curve showing the tangent line at current I=2 to represent dynamic resistance.

Solution:

  1. Given the relation V=I2V = I^2.
  2. To find dynamic resistance rdr_d, we differentiate VV with respect to II: rd=dVdIr_d = \frac{dV}{dI}.
  3. d(I2)dI=2I\frac{d(I^2)}{dI} = 2I.
  4. At I=2 AI = 2 \text{ A}, rd=2×2=4 \Omegar_d = 2 \times 2 = 4 \text{ \Omega}.

Explanation:

For non-linear devices, resistance is not a constant value. The dynamic resistance is the slope of the V−IV-I curve at a specific point, calculated as the derivative of the voltage function with respect to current.