Q25.In the given circuit, the current flowing through the resistance 20 Ω is 0. 3 A, while the ammeter reads 0. 9 A. The value of R1 is _____ Ω .
What This Question Tests
This problem requires applying Ohm's law and Kirchhoff's current law to a parallel circuit to determine an unknown resistance, given currents and one resistance.
Concepts Tested
Formulas Used
V = IR
I_total = I_1 + I_2
📚 NCERT Sections This Tests
3.2 — A Battery Of Emf 10 V And Internal Resistance 3 Ω Is Connected To A
Physics Class 11 · Chapter 3
3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
3.1 — The Storage Battery Of A Car Has An Emf Of 12 V. If The Internal
Physics Class 11 · Chapter 3
3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
3.4 — Ohm’S Law
Physics Class 11 · Chapter 3
3.4 OHM’S LAW A basic law regarding flow of currents was discovered by G.S. Ohm in 1828, long before the physical mechanism responsible for flow of currents was discovered. Imagine a conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor. Then Ohm’s law states that V µ I or, V = R I (3.3) where the constant of proportionality R is called the resistance of the conductor. The SI units of resistance is ohm, and is denoted by the symbol W. The resistance R not only depends on the material of the conductor but also on the dimensions of the conductor. The dependence of R on the dimensions of the conductor can easily be determined as follows. FIGURE 3.2 Consider a conductor satisfying Eq. (3.3) to be in the form of a slab of Illustrating the length l and cross sectional area A [Fig. 3.2(a)]. Imagine placing two such relation R = rl/A for a rectangular slabidentical slabs side by side [Fig. 3.2(b)], so that the length of the of length l and areacombination is 2l. The current flowing through the combination is the of cross-section A. same as that flowing through either of the slabs. If V is the potential difference across the ends of the first slab, then V is also the potential 83difference across the ends of the second slab since the second slab is Reprint 2025-26 Physics identical to the first and the same current I flows through both. The potential difference across the ends of the combination is clearly sum of the potential difference across the two individual slabs and hence equals 2V. The current through the combination is I and the resistance of the combination RC is [from Eq. (3.3)], 2V R C = = 2 R (3.4) I since V/I = R, the resistance of either of the slabs. Thus, doubling the length of a conductor doubles the(1787–1854) resistance. In general, then resistance is proportional to length, R ∝ l (3.5)OHM Georg Simon Ohm (1787– Next, imagine dividing the slab into two by cutting it 1854) German physicist, lengthwise so that the slab can be considered as a professor at Munich. Ohm combination of two identical slabs of length l, but each was led to his law by anSIMON having a cross sectional area of A/2 [Fig. 3.2(c)]. analogy between the For a given voltage V across the slab, if I is the current conduction of heat: the through the entire slab, then clearly the current flowing electric field is analogous to the temperature gradient, through each of the two half-slabs is I/2. Since theGEORG and the electric current is potential difference across the ends of the half-slabs is V, analogous to the heat flow. i.e., the same as across the full slab, the resistance of each of the half-slabs R1 is V V R1 = = 2 = 2 R. (3.6) ( I /2) I Thus, halving the area of the cross-section of a conductor doubles the resistance. In general, then the resistance R is inversely proportional to the cross-sectional area, 1 R ∝ (3.7) A Combining Eqs. (3.5) and (3.7), we have l R ∝ (3.8) A and hence for a given conductor l R = ρ (3.9) A where the constant of proportionality r depends on the material of the conductor but not on its dimensions. r is called resistivity. Using the last equation, Ohm’s law reads I ρl V = I × R = (3.10) A Current per unit area (taken normal to the current), I/A, is called current density and is denoted by j. The SI units of the current density are A/m2. Further, if E is the magnitude of uniform electric field in the conductor whose length is l, then the potential difference V across its 84 ends is El. Using these, the last equation reads Reprint 2025-26 Current Electricity E l = j r l or, E = j r (3.11) The above relation for magnitudes E and j can indeed be cast in a vector form. The current density, (which we have defined as the current through unit area normal to the current) is also directed along E, and is also a vector j (º j E/E). Thus, the last equation can be written as, E = jr (3.12) or, j = s E (3.13) where s º1/r is called the conductivity. Ohm’s law is often stated in an equivalent form, Eq. (3.13) in addition to Eq.(3.3). In the next section, we will try to understand the origin of the Ohm’s law as arising from the characteristics of the drift of electrons.
📋 Question Details
- Chapter
- Current Electricity
- Topic
- Ohm's Law and Parallel Circuits
- Year
- 2024
- Shift
- 29 Jan Shift 2
- Q Number
- Q25
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 3: Current Electricity
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