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Explanation:
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Answer:
Replace the power supply
Explanation:
Usually, Capacitors burn up because of different factors.
- Exceeding operation voltage
- High Inverse voltage if is an electrolytic capacitor.
A tentative answer could be the battery, but usually the batteries damages are because wear out of them, that is, cycles of charging and discharging, a problem that could arise in the battery is related with the charge protection, this circuit cares that the battery only get the charge that it needs, in Li-po batteries is 3.7V, in some laptops is 24 V, if so the battery could explode or leaking acid.
The mother-board is the "brain" and the Random Access Memory (RAM), they consume a lot of energy and usually heat up, but doesn’t produce increasing of voltage and its feed it by voltage regulators.
The only valid option is the power supply because the energy comes from a rectifier (made with diodes) and a voltage regulator that step-down the DC voltage output if by any chance the voltage increase or a diode burns up in the power supply the coupling capacitors or input capacitors in the computer will fail.
Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
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