Subject - Prysics Year 2017 EC 1 A standard coll marised 2 Q is found to have a resistance of 2118 a at 30 OC Calculate the temperature at which the marking is correct. The temperature coefficient of thr renistance of the material of the coll is 0.0042 OC 2. If a wire of resistance \( R \) ohms is stretched to double its length, then its new resistance will become 3. A circuit consists of three resistors in series one of 3 Q , ane of 7 Q and one of 10 Q A battery of e.m.f. 12 V and negliglble internal resistance is connected to the circuit What sine current will it suppty?

**Problem 1: Temperature at Which the Marking is Correct** Given: - Resistance at 30°C = 2118 Ω - Temperature coefficient of resistance = 0.0042/°C We need to find the temperature at which the marking is correct. Using the formula: \[ R = R_0(1 + \alpha(T - T_0)) \] Substituting the given values: \[ 2118 = 2118(1 + 0.0042(T - 30)) \] Simplifying: \[ 1 = 1 + 0.0042(T - 30) \] \[ 0 = 0.0042(T - 30) \] \[ T - 30 = 0 \] \[ T = 30°C \] **Answer:** The temperature at which the marking is correct is **30°C**. **Problem 2: New Resistance When Wire is Stretched** Given: - Initial resistance = \( R \) ohms - New length = 2 times the initial length Since resistance is directly proportional to length: \[ \text{New Resistance} = 2R \] **Answer:** The new resistance of the wire when stretched to double its length is **2R ohms**. **Problem 3: Current Supplied by the Battery** Given: - Resistors in series: 3 Ω, 7 Ω, and 10 Ω - Battery e.m.f. = 12 V - Internal resistance = negligible First, calculate the total resistance of the circuit: \[ R_{\text{total}} = 3 + 7 + 10 = 20 \, \Omega \] Using Ohm's Law: \[ I = \frac{V}{R_{\text{total}}} \] \[ I = \frac{12}{20} = 0.6 \, \text{A} \] **Answer:** The current supplied by the battery is **0.6 A**.