An electric heater of resistance 6 Omega is run 10 min on a line of 120 V- The energy liberated in this period of time is-7-2times 10-3-J14-4times 10-5-J43-2times 10-4-J28-8times 10-4-J

Question

An electric heater of resistance 6 Omega  is run 10 min on a line of 120 V- The energy liberated in this period of time is-7-2times 10-3-J14-4times 10-5-J43-2times 10-4-J28-8times 10-4-J

Toppr Admin · Accepted Answer

By Joule-apos-s Law of heating-E-V2RtE- Energy developedV- Applied voltageR- Resistance of wiret- time-xA0-x21D2-E-12026-xD7-10-xD7-60-x21D2-E-14-4-xD7-105J

An electric heater of resistance $6 Ω$ is run for $10 m i n$ on a line of $120 V$ . The energy liberated in this period of time is:
$7.2 \times 10^{3} J$
$14.4 \times 10^{5} J$
$43.2 \times 10^{4} J$
$28.8 \times 10^{4} J$

By Joule's Law of heating:
$E = \frac{V^{2}}{R} t$
$E$ ; Energy developed
$V$ ; Applied voltage
$R$ ; Resistance of wire
$t$ ; time
$\Rightarrow E = \frac{120^{2}}{6} \times 10 \times 60$
$\Rightarrow E = 14.4 \times 10^{5} J$

An electric heater of resistance 6 Ω is run for 10 min on a line of 120 V. The energy liberated in this period of time is:7.2×103J14.4×105J43.2×104J28.8×104J

By Joule's Law of heating:E=V2RtE; Energy developedV; Applied voltageR; Resistance of wiret; time ⇒E=12026×10×60⇒E=14.4×105J

An electric heater of resistance $6 Ω$ is run for $10 m i n$ on a line of $120 V$ . The energy liberated in this period of time is:
$7.2 \times 10^{3} J$
$14.4 \times 10^{5} J$
$43.2 \times 10^{4} J$
$28.8 \times 10^{4} J$

By Joule's Law of heating:
$E = \frac{V^{2}}{R} t$
$E$ ; Energy developed
$V$ ; Applied voltage
$R$ ; Resistance of wire
$t$ ; time
$\Rightarrow E = \frac{120^{2}}{6} \times 10 \times 60$
$\Rightarrow E = 14.4 \times 10^{5} J$