A resistor of 10kΩ having tolerance 10% is connected in series with another resistor of 20kΩ having tolerance 20%. The tolerance of the combination will be:
A
10%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
13%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
30%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
20%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C30% In series effective resistance =RS=(10kΩ±10%)+(20kΩ±20%)=(30kΩ±30%) ∴ Tolerance of the combination =(±30%)