CUET UG Physics Booster Test – 3 Circuits and Measurement

According to NCERT, the power transmitted is given by: P=VI For a fixed power transmission, I=P/V The power lost in transmission cables of resistance R_c is: P_c=I^2R_c Substituting the value of current: P_c=(P/V)^2R_cP_c=P^2R_c/V^2 Thus, power loss varies inversely as the square of the transmission voltage. This is the reason electrical energy is transmitted at very high voltages. A higher voltage reduces current significantly, resulting in much smaller transmission losses.

In power transmission systems, several important relations are used. The power delivered is: P=VI The current in transmission lines is: I=P/V The power lost due to cable resistance R_c is: P_c=I^2R_c Transformers are used to increase voltage before transmission and reduce it to safe values near consumers. Therefore: • P_c→I^2R_c • P→V×I • I→P/V • Transformer - Lowers voltage at destination This gives the matching 1-c, 2-d, 3-b and 4-a.

When electrodes are immersed in an electrolyte, charge exchange occurs between the electrodes and the electrolyte. As a result, the positive electrode develops a positive potential relative to the electrolyte, represented by V_+. Similarly, the negative electrode develops a non-positive potential represented by -(V_-). Under open-circuit conditions, no current flows through the electrolyte. Therefore, the electrolyte remains at the same potential throughout. The electromotive force of the cell is the total potential difference between the electrodes: ε=V_++V_- which is equivalent to: ε=V_+-(-(V_-)) Thus, all four statements correctly describe electrode potentials and EMF in an electrolytic cell.

When no current flows through an electrolytic cell, the electrolyte remains at the same potential throughout. However, once current begins to flow, charges move through the electrolyte and a potential gradient develops. As a result, different points in the electrolyte may no longer be at the same potential. Therefore, the statement that the electrolyte maintains a uniform potential during current flow is incorrect. Electrodes do exchange charges with the electrolyte, creating electrode potentials. The negative electrode develops a non-positive potential relative to the surrounding electrolyte. For a functioning cell, the electromotive force remains positive. Thus, Option B is the incorrect statement.

Electromotive force (EMF) is defined as the work done by a source per unit charge in moving charge from a lower potential energy state to a higher potential energy state within the source. Mathematically, ε=W/q where W is the work done by the source and q is the charge transported. Although the term contains the word "force", EMF is not a force. It is a potential difference measured in volts. It represents the maximum voltage available from a source under open-circuit conditions. The value of EMF depends on the nature of the source and is independent of the external resistance connected to it. Therefore, Option A correctly defines electromotive force.

The electromotive force (EMF) of an electrolytic cell is the potential difference between its two electrodes under open-circuit conditions. According to NCERT, if V_+is the positive electrode potential and V_-is the magnitude of the negative electrode potential relative to the electrolyte, then: ε=V_++V_- Given: V_+=2.0 VV_-=1.2 V Substituting: ε=2.0+1.2ε=3.2 V Thus, the emf of the cell is 3.2 V. The positive and negative electrode potentials contribute together to the total emf because they represent potential differences relative to the same electrolyte.

Internal resistance is the resistance offered to current flow inside a cell. It arises due to the electrolyte, electrodes and internal construction of the cell. Dry cells generally possess a comparatively higher internal resistance than common electrolytic cells. This is one reason why dry cells cannot supply very large currents continuously. Internal resistance affects the terminal voltage according to: V=ε-Ir Thus, during discharge, the terminal voltage becomes less than the emf. Internal resistance also determines the maximum current that can be supplied by a cell because it opposes current flow. Even under open-circuit conditions, the internal resistance still exists physically; however, no current flows through it. Therefore, Option C is correct.

For a discharging cell, the terminal voltage is related to emf and internal resistance by: V=ε-Ir Given: ε=12.0 VV=10.0 Vr=0.4 Ω Substituting: 10=12-I(0.4)0.4I=2I=2/0.4I=5.0 A Thus, the current flowing through the circuit is 5.0 A. The difference between emf and terminal voltage represents the voltage drop across the internal resistance.

When cells are connected in series, their emfs combine algebraically. If all cells assist one another, their emfs are added directly: ε_(eq)=ε_1+ε_2+ε_3+⋯ If a cell is connected in the opposite direction such that current leaves through its negative terminal, its emf contributes with a negative sign. The equivalent internal resistance of cells connected in series is: r_(eq)=r_1+r_2+r_3+⋯ Therefore, the total internal resistance increases rather than decreases. Series combinations are used when a larger voltage is required. Both the emf and internal resistance increase with the number of cells connected in series. Hence, Statements 1, 2 and 3 are correct.

When cells are connected in series, the same current flows through each cell. Therefore, their internal resistances combine exactly like ordinary resistors connected in series. If the internal resistances are r_1, r_2, r_3, ... then: r_(eq)=r_1+r_2+r_3+⋯ This relation follows directly from Ohm's law and the series combination rule for resistors. As more cells are connected in series, both the total emf and total internal resistance increase. This effect must be considered while designing battery combinations for practical applications. Thus, the equivalent internal resistance is equal to the sum of the individual internal resistances.

For cells connected in parallel, Kirchhoff's Junction Rule gives: I=I_1+I_2 which expresses conservation of charge at a junction. The equivalent internal resistance of two parallel cells is: r_(eq)=r_1r_2/r_1+r_2 For parallel combinations, the quantity e_(eq)/r_(eq) is equal to e_1/r_1+e_2/r_2 which follows from the derivation of equivalent emf. Another important property of parallel circuits is that all branches connected between the same two junctions have the same potential difference. Thus: • I=I_1+I_2→ Charge conservation at junction • r_(eq)→ r_1r_2/r_1+r_2 • e_(eq)/r_(eq)→ e_1/r_1+e_2/r_2 • Potential V→ Same for all branches

For two cells connected in parallel, the equivalent internal resistance is obtained using the parallel resistance formula: 1/r_(eq)=1/r_1+1/r_2 The equivalent emf is: e_(eq)=e_1r_2+e_2r_1/r_1+r_2 These relations can be generalized to any number of cells connected in parallel. When a cell is connected in an orientation opposite to the reference direction, its emf contributes with a negative sign. Therefore, e_2 is replaced by -e_2 in the calculations. All four statements correctly describe the theory of parallel combinations of cells.

Kirchhoff's Junction Rule states that the algebraic sum of currents entering and leaving a junction is zero. ∑I=0 The rule is derived from conservation of charge. In a steady-state circuit, charges cannot continuously accumulate at a junction. If charge accumulation occurred, the current values would change with time and the circuit would not remain in steady state. Therefore, the incoming current must equal the outgoing current. This principle applies not only at branching points but also at any point along a conducting path. Hence, the statement that the rule relies on continuous charge accumulation is incorrect.

Kirchhoff's First Rule, also known as the Junction Rule, follows directly from the conservation of charge. Since current is defined as: I=dq/dt the total rate at which charge enters a junction must equal the total rate at which charge leaves it. Mathematically, ∑I_(in)=∑I_(out) This condition ensures that charge does not accumulate at the junction. The rule is valid for all steady-state circuits, irrespective of whether batteries are present. The rule concerns current conservation and should not be confused with the Loop Rule, which deals with potential differences.

Electric potential is a scalar quantity associated with a particular location in an electric circuit. The potential at one point may differ from the potential at another point due to cells, resistors and other circuit elements. Kirchhoff's Loop Rule is based on the principle that when a charge moves around a closed loop and returns to its starting point, the net change in potential must be zero. ∑ΔV=0 This is possible because electric potential is a location-dependent quantity. As the charge moves through different points in the circuit, it experiences potential rises and potential drops. The algebraic sum of these changes around a complete loop is zero. Therefore, the loop rule fundamentally relies on the concept of potential being associated with location in the circuit.

Kirchhoff's Loop Rule states that the algebraic sum of all potential rises and potential drops around a closed loop is zero. ∑ΔV=0 This follows from the principle of conservation of energy. As a charge moves around a complete circuit and returns to its starting point, its electrical potential energy must return to its original value. Therefore, the net change in potential is zero. Potential rises occur across cells and potential drops occur across resistors. These changes exactly balance one another when summed algebraically around the loop. The rule does not require all currents to be positive, nor does it imply that resistors dissipate zero power. Resistors continue to convert electrical energy into heat according to P=I^2R. Hence, Option C is correct.

For a balanced Wheatstone bridge, R_2/R_1=R_4/R_3 Substituting the given values: 10/100=R_4/601/10=R_4/60 Multiplying both sides by 60: R_4=60×1/10R_4=6Ω Therefore, the bridge will be balanced when the unknown resistance is 6 Ω. At this condition, the galvanometer current becomes zero and the potentials at the galvanometer terminals become equal.

At the null point of a Wheatstone bridge, no current flows through the galvanometer. I_g=0 Since no current passes through the galvanometer branch, the potentials at junctions B and D become equal. Applying Kirchhoff's Junction Rule under this condition gives: I_1=I_3 and I_2=I_4 Applying Kirchhoff's Loop Rule to the appropriate loop gives: -I_1R_1+I_2R_2=0 A similar equation can be written for the other loop. Combining these equations yields the Wheatstone bridge balance condition. Thus, all four statements correctly describe the null point.

When a Wheatstone bridge is balanced, no current flows through the galvanometer. I_g=0 Applying Kirchhoff's Junction Rule and Loop Rule to the bridge under this condition leads to the balance relation: R_2/R_1=R_4/R_3 This equation states that the ratio of resistances in one arm equals the ratio of resistances in the opposite arm. The balance condition forms the basis of resistance measurement using Wheatstone bridges and meter bridges. It allows an unknown resistance to be determined accurately without drawing current through the galvanometer. Therefore, Option A is the correct balance equation.

The meter bridge is a practical application of the Wheatstone bridge principle and is widely used to determine unknown resistances. The expression R_4=R_3(R_2/R_1) is obtained from the bridge balance condition and is used to calculate an unknown resistance. At balance, I_g=0 which means the galvanometer shows no deflection. This is the balance condition of the bridge. The resistances R_1, R_2 and R_3 are usually known quantities used in determining the unknown resistance R_4. Therefore: • Meter bridge → Practical device using bridge principle • Formula → Unknown resistance calculation • I_g=0→ Balance condition • R_1,R_2,R_3→ Known resistances