Class 10 __full__ — Electricity Ncert Solutions

Calculate number of bulbs ($n$) in parallel. For parallel connection: $R_total = \fracRn$ $$44 = \frac4840n$$ $$n = \frac484044 = 110$$ 110 bulbs can be connected.

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1 Answer: (c) 1:4 Explanation: Let resistance of each wire be $R$.

Answer:

The electric iron takes the same current ($7.04 \text A$) from the same source ($220 \text V$). Using Ohm's Law: $$R = \fracVI = \frac2207.04 = 31.25 \ \Omega$$ Resistance of the iron = 31.25 Ω. Current through it = 7.04 A.

When current flows through a purely resistive circuit, the energy is dissipated as heat. This is known as Joule’s Law of Heating. Why Use Electricity NCERT Solutions Class 10? electricity ncert solutions class 10

This is the heart of the chapter. It states that the current flowing through a conductor is directly proportional to the potential difference across its ends, provided temperature remains constant. 4. Resistance and Resistivity

(a) 100 W (b) 75 W (c) 50 W (d) 25 W Answer: (d) 25 W Explanation: We know $P = \fracV^2R$. First, find the resistance of the bulb (which is constant). $$R = \fracV^2P = \frac(220)^2100 = 484 \ \Omega$$ Now, calculate power at 110 V: $$P_new = \frac(V_new)^2R = \frac(110)^2484 = \frac12100484 = 25 \text W$$ Calculate number of bulbs ($n$) in parallel

Answer: Step 1: Calculate total current drawn by the three appliances. Voltage ($V$) = $220 \text V$ Parallel combination resistance ($R_p$): $$\frac1R_p = \frac1100 + \frac150 + \frac1500$$ $$\frac1R_p = \frac5500 + \frac10500 + \frac1500 = \frac16500$$ $$R_p = \frac50016 = 31.25 \ \Omega$$

In parallel, voltage across each resistor is equal to the battery voltage. Voltage across $2 \ \Omega$ resistor = $4 \text V$. Power in $2 \ \Omega$ resistor ($P_2$) = $\fracV^2R = \frac4^22 = \frac162 = 8 \text W$. Answer: The electric iron takes the same current ($7

Answer: (i) Series Circuit (6 V): Total Resistance = $1 + 2 = 3 \ \Omega$. Current in circuit ($I$) = $\frac6 \text V3 \ \Omega = 2 \text A$. Power in $2 \ \Omega$ resistor ($P_1$) = $I^2 R = (2)^2 \times 2 = 4 \times 2 = 8 \text W$.

Mastering Physics: Comprehensive Guide to Electricity NCERT Solutions Class 10